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Sebanyak 100 mL larutan 0,2 M dicampur dengan 100 mL larutan 0,2 M. Jika , pH larutan setelah dicampur adalah ....

Pertanyaan

Sebanyak 100 mL larutan C H subscript 3 C O O H 0,2 M dicampur dengan 100 mL larutan Na O H 0,2 M. Jika K subscript a C H subscript 3 C O O H equals 1 cross times 10 to the power of negative sign 5 end exponent, pH larutan setelah dicampur adalah ....space

  1. 2space

  2. 4space

  3. 5space

  4. 6space

  5. 9space

Pembahasan Soal:

Reaksi yang terjadi adalah


Karena asam dan basa habis bereaksi, maka terbentuk campuran garam terhidrolisis basa.

table attributes columnalign right center left columnspacing 0px end attributes row cell open square brackets O H to the power of minus sign close square brackets end cell equals cell square root of K subscript w over K subscript a cross times open square brackets C H subscript 3 C O O Na close square brackets end root end cell row cell open square brackets O H to the power of minus sign close square brackets end cell equals cell square root of 10 to the power of negative sign 14 end exponent over 10 to the power of negative sign 5 end exponent cross times left parenthesis bevelled fraction numerator 20 space mmol over denominator 200 space mL end fraction right parenthesis end root end cell row cell open square brackets O H to the power of minus sign close square brackets end cell equals cell square root of 10 to the power of negative sign 9 end exponent cross times 0 comma 1 end root end cell row cell open square brackets O H to the power of minus sign close square brackets end cell equals cell 10 to the power of negative sign 5 end exponent end cell row pOH equals cell negative sign log space open square brackets O H to the power of minus sign close square brackets end cell row pOH equals cell negative sign log space 10 to the power of negative sign 5 end exponent end cell row pOH equals 5 row pH equals cell 14 minus sign pOH end cell row pH equals cell 14 minus sign 5 end cell row bold pH bold equals bold 9 end table


Jadi, jawaban yang tepat adalah E.space

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

B. Rohmawati

Mahasiswa/Alumni Universitas Negeri Semarang

Terakhir diupdate 02 Mei 2021

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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Pertanyaan yang serupa

Sebuah wadah berisi 100 mL larutan asam fluorida 0,01 M dengan pH = 3. Jika ke dalam wadah tersebut ditambahkan 100 mL larutan kalium hidroksida 0,01 M, pH setelah penambahan sebesar ....

Pembahasan Soal:

Larutan asam fluorida merupakan senyawa asam lemah, di reaksikan dengan kalium hidroksida yang bersifat basa kuat. pH campurannya yaitu:

  1. Menghitung mol maisng-masing zat

    begin mathsize 14px style mol space H F equals 100 space ml cross times 0 comma 01 space M equals 1 space mmol mol space K O H equals 100 space ml cross times 0 comma 01 space M equals 1 space mmol end style 
     
  2. Membuat mrs

    begin mathsize 14px style space space space space space space H F and K O H yields K F and H subscript 2 O m space colon space space 1 space space space space space space space space 1 space space space space space space space space minus sign space space space space space minus sign bottom enclose r space space space colon space space 1 space space space space space space space space 1 space space space space space space space space space 1 space space space space space space space 1 end enclose space space s space space colon space space minus sign space space space space space space minus sign space space space space space space space space space 1 space space space space space space space 1 end style 

    Karena yang bersisa hanya garam begin mathsize 14px style open parentheses K F close parentheses end style yang bersifat basa, maka perhitungan pH menggunakan rumus hidrolisis garam basa.
     
  3. Menghitung pH garam basa

    begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell open square brackets g close square brackets end cell rightwards double arrow cell M equals n over Vtotal end cell row blank equals cell fraction numerator 1 space mmol over denominator 200 space ml end fraction end cell row blank equals cell 0 comma 005 space M end cell row blank equals cell 5 cross times 10 to the power of negative sign 3 end exponent space M end cell end table end style 

    begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell open square brackets O H to the power of minus sign close square brackets end cell equals cell square root of Kw over Ka cross times open square brackets g close square brackets end root end cell row blank equals cell square root of 10 to the power of negative sign 14 end exponent over 10 to the power of negative sign 4 end exponent cross times 5 cross times 10 to the power of negative sign 3 end exponent end root end cell row blank equals cell square root of 5 cross times 10 to the power of negative sign 13 end exponent end root end cell row blank equals cell 2 comma 2 cross times 10 to the power of negative sign 6 comma 5 end exponent end cell end table end style 

    begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row pOH equals cell negative sign log space open square brackets O H to the power of minus sign close square brackets end cell row blank equals cell negative sign log space 2 comma 2 cross times 10 to the power of negative sign 6 comma 5 end exponent end cell row blank equals cell 6 comma 5 minus sign log space 2 comma 2 end cell end table end style 

    begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row pH equals cell 14 minus sign pOH end cell row blank equals cell 14 minus sign open parentheses 6 comma 5 minus sign log space 2 comma 2 close parentheses end cell row blank equals cell 7 comma 5 plus log space 2 comma 2 end cell end table end style 

Jadi, pH campuran tersebut adalah 7,5 + log 2,2.space 

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Roboguru

Kebersihan merupakan sebagian dari iman. Baju yang kotor dapat dibersihkan dengan detergen. Detergen mengandung bayclin. Garam yang terdapat di dalam bayclin yaitu natrium hipoklorit (NaOCI). Sebanyak...

Pembahasan Soal:

Perhitungan dari proses ini yaitu:
 

table attributes columnalign right center left columnspacing 0px end attributes row pOH equals cell 14 minus sign 8 end cell row blank equals 6 row cell open square brackets O H to the power of minus sign close square brackets end cell equals cell 10 to the power of negative sign 6 end exponent end cell row blank blank blank row cell open square brackets O H to the power of minus sign close square brackets end cell equals cell square root of Kw over Ka cross times open square brackets G close square brackets end root end cell row cell 10 to the power of negative sign 6 end exponent end cell equals cell square root of fraction numerator 10 to the power of negative sign 14 end exponent over denominator 2 comma 9 cross times 10 to the power of negative sign 8 end exponent end fraction cross times open square brackets G close square brackets end root end cell row cell open square brackets G close square brackets end cell equals cell 2 comma 9 cross times 10 to the power of negative sign 6 end exponent space M end cell row blank blank blank row cell Na O Cl end cell rightwards arrow cell Na to the power of plus sign and O Cl to the power of minus sign end cell row cell 2 comma 9 cross times 10 to the power of negative sign 6 end exponent end cell rightwards arrow cell 2 comma 9 cross times 10 to the power of negative sign 6 end exponent plus 2 comma 9 cross times 10 to the power of negative sign 6 end exponent end cell row blank blank blank row cell n space Na O Cl end cell equals cell m over Mr end cell row blank equals cell fraction numerator 20 cross times 10 to the power of negative sign 3 end exponent space g over denominator 74 comma 5 space g forward slash mol end fraction end cell row blank equals cell 2 comma 68 cross times 10 to the power of negative sign 4 end exponent space mol end cell row blank blank blank row cell open square brackets Na O Cl close square brackets end cell equals cell n over V end cell row cell 2 comma 9 cross times 10 to the power of negative sign 6 end exponent end cell equals cell fraction numerator 2 comma 68 cross times 10 to the power of negative sign 4 end exponent space mol over denominator V end fraction end cell row V equals cell 92 comma 4 space L equals 92400 space ml end cell row blank blank blank row V equals cell V space H O Cl and V space Na O H end cell row 92400 equals cell 25 plus V space Na O H end cell row cell V space Na O H end cell equals cell 92375 space ml end cell row blank blank blank end table 
 

Jadi, dapat disimpulkan jawaban yang tepat yaitu 92375 ml.space

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Roboguru

Sebanyak  kristal  ditambahkan ke dalam  larutan . pH larutan yang teriadi jika diketahui , dan  adalah...

Pembahasan Soal:

Hidrolisis garam adalah reaksi antara komponen garam yang berasal dari asam atau basa lemah dengan air. Untuk menghitung pH larutan tersebut dapat dilakukanspacetahapan:

  • Menghitung mol K O H

table attributes columnalign right center left columnspacing 0px end attributes row cell mol space K O H end cell equals cell massa over Mr end cell row blank equals cell fraction numerator 0 comma 28 over denominator 56 end fraction end cell row blank equals cell 0 comma 005 space mol end cell row blank equals cell 5 space mmol end cell end table 
 

  • Menghitung mol C H subscript 3 C O O H 

table attributes columnalign right center left columnspacing 0px end attributes row cell mol space C H subscript 3 C O O H end cell equals cell M cross times V end cell row blank equals cell 0 comma 1 space M cross times 50 space mL end cell row blank equals cell 5 space mmol end cell end table 

  • Masukkan ke tabel MRS
     

 

 

  • Garam C H subscript 3 C O O K berasal dari asam lemah dan basa kuat mengalami hidrolisis parsial dan larutannya bersifat basa.
  • Menghitung open square brackets O H to the power of minus sign close square brackets space C H subscript 3 C O O K

table attributes columnalign right center left columnspacing 0px end attributes row cell open square brackets O H to the power of minus sign close square brackets end cell equals cell square root of K subscript w over K subscript a cross times M space C H subscript 3 C O O Na cross times valensi end root end cell row blank equals cell square root of 10 to the power of negative sign 14 end exponent over 10 to the power of negative sign 5 end exponent cross times fraction numerator 5 space mmol over denominator 50 space mL end fraction cross times 1 end root end cell row blank equals cell square root of 10 to the power of negative sign 9 end exponent cross times 10 to the power of negative sign 1 end exponent cross times 1 end root end cell row blank equals cell square root of 10 to the power of negative sign 10 end exponent end root end cell row blank equals cell 10 to the power of negative sign 5 end exponent end cell end table

 

  • Menghitung pOH
     

table attributes columnalign right center left columnspacing 0px end attributes row pOH equals cell negative sign log open square brackets O H to the power of minus sign close square brackets end cell row blank equals cell negative sign log space 10 to the power of negative sign 5 end exponent end cell row blank equals 5 end table
 

  • Menghitung pH
     

table attributes columnalign right center left columnspacing 0px end attributes row pH equals cell 14 minus sign pOH end cell row blank equals cell 14 minus sign 5 end cell row blank equals 9 end table


Jadi, jawaban yang benar adalahspaceD.

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Roboguru

Larutan NaX 0,1 M terhidrolisis 10%. Hitunglah tetapan hidrolisis garam tersebut dan pHlarutannya!

Pembahasan Soal:

Hidrolisis adalah reaksi perurauian garam oleh air atau reaksi antarab ion-ion garam dengan air. Garam yang kationnya berasal dari basa kuat dan anionnya berasal dari asam lemah seperti garam NaX mempunyai sifatspacebasa.

Menghitung nilai tetapan hidrolisis garam

table attributes columnalign right center left columnspacing 0px end attributes row alpha equals cell square root of Kh over M end root end cell row cell alpha squared end cell equals cell Kh over M end cell row cell left parenthesis 10 percent sign right parenthesis squared end cell equals cell fraction numerator Kh over denominator 0 comma 1 end fraction space end cell row cell 0 comma 01 end cell equals cell fraction numerator Kh over denominator 0 comma 1 end fraction end cell row Kh equals cell 0 comma 001 end cell end table


Menghitung open square brackets O H to the power of minus sign close square brackets

NaX space space space yields Na to the power of plus sign space plus space X to the power of minus sign 0 comma 1 space M space space space space 0 comma 1 space M space space 0 comma 1 space M  

table attributes columnalign right center left columnspacing 0px end attributes row cell open square brackets O H to the power of minus sign close square brackets end cell equals cell square root of Kw over Ka cross times left square bracket anion space NaX right square bracket end root end cell row blank equals cell square root of Kh cross times left square bracket anion space NaX right square bracket end root end cell row blank equals cell square root of 0 comma 001 cross times 0 comma 1 end root end cell row blank equals cell square root of 0 comma 0001 end root end cell row blank equals cell 0 comma 01 end cell row blank equals cell 1 cross times 10 to the power of negative sign 2 end exponent end cell row blank blank blank end table 

Menghitung pOH larutan

table attributes columnalign right center left columnspacing 0px end attributes row pOH equals cell negative sign log open square brackets O H to the power of minus sign close square brackets end cell row blank equals cell negative sign log space 10 to the power of negative sign 2 end exponent end cell row blank equals 2 end table
 

Menghitung pH larutan

table attributes columnalign right center left columnspacing 0px end attributes row pH equals cell 14 minus sign pOH end cell row blank equals cell 14 minus sign 2 end cell row blank equals 12 end table
 

Jadi, tetapan hidrolisis garam tersebut adalah 0,001 dan pH larutannya adalahspace12.

0

Roboguru

Jika 100 mL larutan  0,2 M () direaksikan dengan 100 mL larutan  0,1 M, pH larutan garam yang terbentuk sebesar ....

Pembahasan Soal:

Tahap.1 Menentukan begin mathsize 14px style open square brackets C H subscript 3 C O O to the power of minus sign close square brackets end style

Berdasarkan informasi yang diberikan pada soal, reaksi yang terjadi antara 100 mL larutan begin mathsize 14px style C H subscript 3 C O O H end style 0,2 M dan 100 mL larutan begin mathsize 14px style Ba open parentheses O H close parentheses subscript 2 end style 0,1 M adalah sebagai berikut.

begin mathsize 14px style n space C H subscript 3 C O O H equals 0 comma 2 space mol space L to the power of negative sign 1 end exponent cross times 100 space mL n space C H subscript 3 C O O H equals 20 space mmol space space space space n space Ba open parentheses O H close parentheses subscript 2 equals 0 comma 1 space mol space L to the power of negative sign 1 end exponent cross times 100 space mL space space space space n space Ba open parentheses O H close parentheses subscript 2 equals 10 space mmol end style 


 

begin mathsize 14px style open parentheses C H subscript 3 C O O close parentheses subscript 2 Ba yields 2 C H subscript 3 C O O to the power of minus sign and Ba to the power of plus sign end style 

begin mathsize 14px style n space C H subscript 3 C O O to the power of minus sign equals 2 over 1 cross times n space open parentheses C H subscript 3 C O O close parentheses subscript 2 Ba n space C H subscript 3 C O O to the power of minus sign equals 2 cross times 10 space mmol n space C H subscript 3 C O O to the power of minus sign equals 20 space mmol end style 

begin mathsize 14px style open square brackets C H subscript 3 C O O to the power of minus sign close square brackets equals fraction numerator n space C H subscript 3 C O O to the power of minus sign over denominator V space total end fraction open square brackets C H subscript 3 C O O to the power of minus sign close square brackets equals fraction numerator 20 space mmol over denominator 200 space mL end fraction open square brackets C H subscript 3 C O O to the power of minus sign close square brackets equals 0 comma 1 space M end style 

Tahap 2. Menentukan pH larutan

Hasil dari reaksi adalah garam begin mathsize 14px style open parentheses C H subscript 3 C O O close parentheses subscript 2 Ba end style yang mengalami hidrolisis parsial dan menghasilkan ion begin mathsize 14px style O H to the power of minus sign end style

begin mathsize 14px style C H subscript 3 C O O to the power of minus sign and H subscript 2 O yields C H subscript 3 C O O H and O H to the power of minus sign end style 

begin mathsize 14px style open square brackets OH to the power of minus close square brackets equals square root of straight K subscript straight w over straight K subscript straight a cross times open square brackets CH subscript 3 COO to the power of minus close square brackets end root open square brackets OH to the power of minus close square brackets equals square root of 10 to the power of negative 14 end exponent over 10 to the power of negative 5 end exponent cross times 0 comma 1 end root open square brackets OH to the power of minus close square brackets equals square root of 10 to the power of negative 9 end exponent cross times 10 to the power of negative 1 end exponent end root open square brackets OH to the power of minus close square brackets equals square root of 10 to the power of negative 10 end exponent end root open square brackets OH to the power of minus close square brackets equals 10 to the power of negative 5 end exponent end style 

begin mathsize 14px style pH equals 14 minus pOH pH equals 14 minus left parenthesis negative log space open square brackets OH to the power of minus close square brackets right parenthesis pH equals 14 minus left parenthesis negative log space 10 to the power of negative 5 end exponent right parenthesis pH equals 14 minus 5 pH equals 9 end style 

Jadi, jawaban yang benar adalah C.undefined 

0

Roboguru

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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