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Sebanyak 100 mL larutan  0,15 M dicampurkan dengan 100 mL larutan HCI 0,10 M (), tentukan pH campuran!

Pertanyaan

Sebanyak 100 mL larutan N H subscript 4 O H 0,15 M dicampurkan dengan 100 mL larutan HCI 0,10 M (K subscript b space N H subscript 4 O H space equals space 10 to the power of negative sign 5 end exponent), tentukan pH campuran!

Pembahasan Soal:

Untuk menentukan jenis campuran dan pH campuran asam basa, kita tuliskan persamaan reaksi mrs terlebih dahulu.

table attributes columnalign right center left columnspacing 0px end attributes row cell mol space N H subscript 4 O H end cell equals cell M cross times V end cell row blank equals cell 0 comma 15 space M cross times 100 space mL end cell row blank equals cell 15 space mmol end cell row cell mol space H Cl end cell equals cell M cross times V end cell row blank equals cell 0 comma 10 space M cross times 100 space mL end cell row blank equals cell 10 space mmol end cell end table 

Perhatikan bahwa di akhir reaksi terdapat basa lemah N H subscript 4 O H dan garam N H subscript 4 Cl yang tersisa yang menandakan campuran ini membentuk sistem penyangga basa. Dengan demikian, pHnya dapat ditentukan sebagai berikut:

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 K subscript b cross times fraction numerator open square brackets N H subscript 4 O H close square brackets over denominator open square brackets N H subscript 4 Cl close square brackets end fraction end cell row blank equals cell 10 to the power of negative sign 5 end exponent cross times fraction numerator fraction numerator 5 space mmol over denominator up diagonal strike 200 space mL end strike end fraction over denominator fraction numerator 10 space mmol over denominator up diagonal strike 200 space mL end strike end fraction end fraction end cell row blank equals cell 10 to the power of negative sign 5 end exponent cross times 0 comma 5 end cell row blank equals cell 5 cross times 10 to the power of negative sign 6 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 blank equals cell negative sign log space 5 cross times 10 to the power of negative sign 6 end exponent end cell row blank equals cell 6 minus sign log space 5 end cell row cell pH and pOH end cell equals 14 row pH equals cell 14 minus sign left parenthesis 6 minus sign log space 5 right parenthesis end cell row pH equals cell 8 plus log space 5 end cell end table 

Jadi, pH campuran tersebut adalah 8 + log 5.

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

Q. 'Ainillana

Mahasiswa/Alumni Universitas Negeri Yogyakarta

Terakhir diupdate 07 Juni 2021

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Pertanyaan yang serupa

Diketahui tetapan ionisasi asam asetat dan amonia adalah . Tersedia larutan  masing-masing 0,1 M. Larutan buffer dengan pH=9 dapat dibuat dengan mencampurkan ...

Pembahasan Soal:

Larutan penyangga dengan pH 9 adalah larutan yang bersifat basa yang berasal dari basa lemah dengan asam konjugasinya, pada akhir reaksi harus tersisa basa lemah dan garamnya. 

Pada pilihan jawaban di atas yang menghasilkan sisa basa lemah di akhir reaksi adalah campuran H Cl space dan space N H subscript 3 dengan volume 1:2. 

Mencari mol N H subscript bold 3 bold space bold dan bold space H Cl

table attributes columnalign right center left columnspacing 0px end attributes row cell mol space N H subscript 3 end cell equals cell V space N H subscript 3 cross times M space N H subscript 3 end cell row cell mol space N H subscript 3 end cell equals cell 2 V cross times 0 comma 1 space M end cell row cell mol space N H subscript 3 end cell equals cell 0 comma 2 V space mmol end cell row blank blank blank row cell mol space H Cl end cell equals cell V space H Cl cross times M space H Cl end cell row cell mol space H Cl end cell equals cell 1 V cross times 0 comma 1 space M end cell row cell mol space H Cl end cell equals cell 0 comma 1 V space mmol end cell end table 
 

Reaksi antara N H subscript bold 3 bold space bold dan bold space H Cl
 


 

Menentukan pH campuran

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 K subscript b cross times fraction numerator mol space basa space lemah over denominator mol space asam space konjugasi end fraction end cell row cell open square brackets O H to the power of minus sign close square brackets end cell equals cell K subscript b cross times fraction numerator mol space N H subscript 4 O H over denominator mol space N H subscript 4 Cl end fraction 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 cross times fraction numerator 0 comma 1 V space mmol over denominator 0 comma 1 V space mmol end fraction 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 space M end cell row blank blank blank 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 open square brackets 10 to the power of negative sign 5 end exponent space M close square brackets end cell row pOH equals 5 row blank blank blank row pH equals cell 14 minus sign pOH end cell row pH equals cell 14 minus sign 5 end cell row pH equals 9 end table 

Jadi yang menghasilkan sisa basa lemah di akhir reaksi adalah campuran H Cl bold space bold dan bold space N H subscript bold 3 dengan volume 1:2. 

Oleh karena itu jawaban yang benar adalah B.space 

Roboguru

Tentukan pH larutan penyangga yang dibuat dengan mencampurkan 20 ml larutan   0,1 M  dan larutan KOH 0,1 M

Pembahasan Soal:

Roboguru

Campuran yang termasuk larutan penyangga adalah ...

Pembahasan Soal:

Syarat terbentuknya larutan penyangga melalui reaksi pencampuran asam dan basa adalah:

  • Salah satu pereaksinya adalah asam lemah / basa lemah
  • Jumlah mol pereaksi asam/basa lemah harus lebih besar dari jumlah mol pereaksi asam/basa kuat. Dengan kata lain, pereaksi pembatasnya adalah asam/basa kuat.
  • Pada akhir reaksi tersisa asam lemah / basa lemah dengan garamnya.
     
  1. 100 space mL space Na O H thin space 0 comma 1 space M plus 50 space mL space C H subscript 3 C O O H space 0 comma 1 space M space space space

    space space space space space space space space space space space Na O H space space space space space space plus space space C H subscript 3 C O O H space space space rightwards arrow space space C H subscript 3 C O O Na space plus space space H subscript 2 O space space space 100 space mL cross times 0 comma 1 space M space space space 50 space mL cross times 0 comma 1 space M M space space space space space space 10 space mmol space space space space space space space space space space space space space 5 space mmol space space space space space space space space space space space space space space space space space space space minus sign space space space space space space space space space space space space space space space space minus sign bottom enclose R space space space space space minus sign 5 space mmol space space space space space space space space space space minus sign 5 space mmol space space space space space space space space space space space space plus 5 space mmol space space space space space plus 5 space mmol end enclose S space space space space space space space space 5 space mmol space space space space space space space space space space space space space space space space space space space minus sign space space space space space space space space space space space space space space space space space space 5 space mmol space space space space space space space space space 5 space mmol space  

     
  2. 100 space mL space Na O H thin space 0 comma 1 space M plus 100 space mL space C H subscript 3 C O O H space 0 comma 1 space M space space space

    space space space space space space space space space space space Na O H space space space space space space plus space space C H subscript 3 C O O H space space space space space rightwards arrow space space C H subscript 3 C O O Na space plus space space H subscript 2 O space space space 100 space mL cross times 0 comma 1 space M space space space 100 space mL cross times 0 comma 1 space M M space space space space space space 10 space mmol space space space space space space space space space space space space space 10 space mmol space space space space space space space space space space space space space space space space space space space minus sign space space space space space space space space space space space space space space space space minus sign bottom enclose R space space space minus sign 10 space mmol space space space space space space space space space space minus sign 10 space mmol space space space space space space space space space space plus 10 space mmol space space space plus 10 space mmol end enclose S space space space space space space space space space space space minus sign space space space space space space space space space space space space space space space space space space space space space space space space minus sign space space space space space space space space space space space space space space space space space space space 10 space mmol space space space space space space 10 space mmol space     

     
  3. 100 space mL space N H subscript 4 O H thin space 0 comma 1 space M plus 50 space mL space H Cl space 0 comma 1 space M space space space

    space space space space space space space space space space space N H subscript 4 O H space space space space space space plus space space space space space space space H Cl space space space space space space space space space space rightwards arrow space space N H subscript 4 Cl space space space space space plus space space space space H subscript 2 O space space space 100 space mL cross times 0 comma 1 space M space space space 50 space mL cross times 0 comma 1 space M M space space space space space space 10 space mmol space space space space space space space space space space space space space 5 space mmol space space space space space space space space space space space space space space space space space space space minus sign space space space space space space space space space space space space space space space space minus sign bottom enclose R space space space space space minus sign 5 space mmol space space space space space space space space space space minus sign 5 space mmol space space space space space space space space space space space space plus 5 space mmol space space space space space plus 5 space mmol end enclose S space space space space space space space space 5 space mmol space space space space space space space space space space space space space space space space space space space minus sign space space space space space space space space space space space space space space space space space space 5 space mmol space space space space space space space space space 5 space mmol space  

     
  4. 100 space mL space N H subscript 4 O H thin space 0 comma 1 space M plus 100 space mL space H Cl space 0 comma 1 space M space space space

    space space space space space space space space space space N H subscript 4 O H space space space space space space plus space space space space space space space space H Cl space space space space space space space space space rightwards arrow space space space space space N H subscript 4 Cl space space space space plus space space H subscript 2 O space space space 100 space mL cross times 0 comma 1 space M space space space 100 space mL cross times 0 comma 1 space M M space space space space space space 10 space mmol space space space space space space space space space space space space space 10 space mmol space space space space space space space space space space space space space space space space space space space minus sign space space space space space space space space space space space space space space space space minus sign bottom enclose R space space space minus sign 10 space mmol space space space space space space space space space space minus sign 10 space mmol space space space space space space space space space space plus 10 space mmol space space space plus 10 space mmol end enclose S space space space space space space space space space space space minus sign space space space space space space space space space space space space space space space space space space space space space space space space minus sign space space space space space space space space space space space space space space space space space space space 10 space mmol space space space space space space 10 space mmol space 

Dari keempat reaksi asam basa tersebut, yang hasil akhir reaksinya tersisa basa lemah dan garamnya adalah reaksi 100 space mL space N H subscript 4 O H thin space 0 comma 1 space M plus 50 space mL space H Cl space 0 comma 1 space M. Untuk reaksi antara NaOH dan HCl bukanlah reaksi pembentukan buffer karena keduanya merupakan basa dan asam kuat.


Jadi, jawabannya adalah C.space space space

Roboguru

Suatu campuran , dengan  mempuyai perbandingan mol = 1:9, jika . Tentukan pH campuran!

Pembahasan Soal:

Campuran yang terdiri dari basa lemah dan garam akan membentuk larutan penyangga maka untuk menentukan pH campurannya digunakan pH larutan penyangga


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 K subscript b cross times fraction numerator mol space asam over denominator mol space basa end fraction end cell row blank equals cell 10 to the power of negative sign 5 end exponent cross times 1 over 9 end cell row blank equals cell 1 comma 1 cross times 10 to the power of negative sign 6 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 blank equals cell negative sign log space 1 comma 1 cross times 10 to the power of negative sign 6 end exponent end cell row blank equals cell 6 minus sign log space 1 comma 1 end cell row pH equals cell 14 minus sign left parenthesis 6 minus sign log space 1 comma 1 right parenthesis end cell row blank equals cell 8 plus log space 1 comma 1 end cell end table end style


Jadi, pH campuran tersebut adalah 8+log 1,1.space 

Roboguru

Berapakah massa  yang harus ditambahkan ke dalam larutan  apabila Anda ingin membuat larutan penyangga dengan pH = 4 dari 100 mL larutan tersebut?

Pembahasan Soal:

Larutan penyangga tersebut adalah larutan penyangga bersifat asam karena reaksinya mengeluarkan ion begin mathsize 14px style H to the power of plus sign end style.

begin mathsize 14px style C H subscript 3 C O O H left parenthesis italic a italic q right parenthesis equilibrium C H subscript 3 C O O to the power of minus sign left parenthesis italic a italic q right parenthesis plus H to the power of plus sign left parenthesis italic a italic q right parenthesis C H subscript 3 C O O Na left parenthesis italic a italic q right parenthesis yields C H subscript 3 C O O to the power of minus sign left parenthesis italic a italic q right parenthesis plus Na to the power of plus sign left parenthesis italic a italic q right parenthesis end style 

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell mmol space C H subscript 3 C O O H end cell equals cell V space C H subscript 3 C O O H cross times space M space C H subscript 3 C O O H end cell row cell mmol space C H subscript 3 C O O H end cell equals cell 100 space mL cross times 0 comma 1 space mmol forward slash mL end cell row cell mmol space C H subscript 3 C O O H end cell equals cell 10 space mmol end cell row blank blank blank row pH equals 4 row cell open square brackets H to the power of plus sign close square brackets end cell equals cell 10 to the power of negative sign 4 end exponent space M end cell row blank blank blank row cell open square brackets H to the power of plus sign close square brackets end cell equals cell K subscript italic a cross times fraction numerator mmol space asam space lemah over denominator mol space garam end fraction end cell row cell open square brackets H to the power of plus sign close square brackets end cell equals cell K subscript italic a cross times fraction numerator mmol space C H subscript 3 C O O H over denominator mmol space C H subscript 3 C O O Na end fraction end cell row blank blank blank row cell 10 to the power of negative sign 4 end exponent end cell equals cell 1 comma 8 cross times 10 to the power of negative sign 5 end exponent cross times fraction numerator 10 space mmol over denominator mmol space C H subscript 3 C O O Na end fraction end cell row cell mmol space C H subscript 3 C O O Na end cell equals cell fraction numerator 1 comma 8 cross times 10 to the power of negative sign 5 end exponent cross times 10 space mmol over denominator 10 to the power of negative sign 4 end exponent end fraction end cell row cell mmol space C H subscript 3 C O O Na end cell equals cell 1 comma 8 space mmol end cell row cell mmol space C H subscript 3 C O O Na end cell equals cell fraction numerator italic g space C H subscript 3 C O O Na over denominator M subscript r space C H 3 C O O Na end fraction x 1000 end cell row cell 1 comma 8 space mmol end cell equals cell fraction numerator italic g space C H subscript 3 C O O Na over denominator 82 space gram space mol to the power of negative sign 1 end exponent end fraction x 1000 end cell row cell italic g space C H subscript 3 C O O Na end cell equals cell fraction numerator 1 comma 8 space mmol cross times 82 space gram space mol to the power of negative sign 1 end exponent over denominator 1000 end fraction end cell row cell italic g space C H subscript 3 C O O Na end cell equals cell 0 comma 1476 space gram end cell row blank blank blank end table end style 

Jadi massa begin mathsize 14px style C H subscript bold 3 C O O Na end style yang harus ditambahkan adalah 0,1476 gram.space 

Roboguru

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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