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Banyaknya garam natrium sulfit  yang harus dilarutkan dalam 500 ml air agar diperoleh pH = 10 adalah ....  dan )

Pertanyaan

Banyaknya garam natrium sulfit open parentheses Na subscript 2 S O subscript 3 close parentheses yang harus dilarutkan dalam 500 ml air agar diperoleh pH = 10 adalah .... left parenthesis K subscript a space H subscript 2 S O subscript 3 equals 1 comma 0 cross times 10 to the power of negative sign 7 end exponent dan M subscript r space Na subscript 2 S O subscript 3 equals 126 space g space mol to the power of negative sign 1 end exponent)

  1. 0,78 gram

  2. 1,56 gram

  3. 3,15 gram

  4. 6,30 gram

  5. 12,6 gram

Pembahasan Soal:

Garam natrium sulfit open parentheses Na subscript 2 S O subscript 3 close parentheses merupakan garam yang mengalami hidrolisis parsial dan bersifat basa. Hal ini disebabkan komponen asam lemah dari garam Na subscript 2 S O subscript 3, yaitu S O subscript 3 to the power of 2 minus sign end exponent bereaksi dengan air melepas ion O H to the power of minus sign. Jika pH garam adalah 10, maka pOH-nya adalah 14 minus sign 10 equals 4. Untuk mengetahui massa garam, kita cari M garam terlebih dahulu.

table attributes columnalign right center left columnspacing 0px end attributes row pOH equals 4 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 4 equals cell negative sign log space log space open square brackets O H to the power of minus sign close square brackets 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 4 end exponent 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 K subscript w over K subscript a cross times left square bracket S O subscript 3 to the power of 2 minus sign end exponent right square bracket end root end cell row cell 10 to the power of negative sign 4 end exponent end cell equals cell square root of fraction numerator 10 to the power of negative sign 14 end exponent over denominator 1 cross times 10 to the power of negative sign 7 end exponent end fraction cross times left square bracket S O subscript 3 to the power of 2 minus sign end exponent right square bracket end root left parenthesis kuadratkan space kedua space ruas right parenthesis end cell row cell 10 to the power of negative sign 8 end exponent end cell equals cell 10 to the power of negative sign 7 end exponent cross times left square bracket S O subscript 3 to the power of 2 minus sign end exponent right square bracket end cell row cell left square bracket S O subscript 3 to the power of 2 minus sign end exponent right square bracket end cell equals cell 10 to the power of negative sign 1 end exponent end cell end table 

Karena konsentrasi S O subscript 3 to the power of 2 minus sign end exponent = konsentrasi (M) garam, maka:

table attributes columnalign right center left columnspacing 0px end attributes row M equals cell massa over M subscript r cross times 1000 over V end cell row cell 0 comma 1 end cell equals cell massa over 126 cross times fraction numerator 1000 over denominator 500 space mL end fraction end cell row cell 0 comma 1 end cell equals cell fraction numerator massa cross times 2 over denominator 126 end fraction end cell row massa equals cell 6 comma 3 space gram end cell end table 

Sehingga massa garam Na subscript 2 S O subscript 3 adalah 6,3 gram.

Jadi, jawaban yang benar adalah D.

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

Q. 'Ainillana

Mahasiswa/Alumni Universitas Negeri Yogyakarta

Terakhir diupdate 21 September 2021

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Larutan begin mathsize 14px style Na subscript 2 C O subscript 3 end style merupakan larutan garam yang terdiri dari sisa basa kuat dan sisa asam lemah yang dalam air mengalami hidrolisis parsial, dimana sisa asam lemah bereaksi dengan air melepas ion begin mathsize 14px style O H to the power of minus sign end style. Jika diketahui 2 nilai begin mathsize 14px style K subscript a end style, seharusnya dicari konsentrasi begin mathsize 14px style O H to the power of minus sign end style untuk masing-masing begin mathsize 14px style K subscript a end style kemudian dijumlahkan. Namun biasanya karena nilai salah satu begin mathsize 14px style K subscript a end style sangat kecil, maka dapat diabaikan. Sehingga, untuk garam terhidrolisis, nilai begin mathsize 14px style K subscript a end style yang digunakan adalah begin mathsize 14px style K subscript a end style yang kecil, yaitu begin mathsize 14px style K subscript a 2 end subscript 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 K subscript w over K subscript a 2 end subscript cross times left square bracket C O subscript 3 to the power of 2 minus sign end exponent right square bracket end root end cell row blank equals cell square root of fraction numerator 10 to the power of negative sign 14 end exponent over denominator 6 cross times 10 to the power of negative sign 11 end exponent end fraction cross times 0 comma 1 end root end cell row blank equals cell square root of 10 to the power of negative sign 4 end exponent over 6 end root end cell row blank equals cell 0 comma 408 cross times 10 to the power of negative sign 2 end exponent end cell row blank equals cell 4 comma 08 cross times 10 to the power of negative sign 3 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 4 comma 08 cross times 10 to the power of negative sign 3 end exponent end cell row blank equals cell 3 minus sign log space 4 comma 08 end cell row cell pH and pOH end cell equals 14 row pH equals cell 14 minus sign left parenthesis 3 minus sign log space 4 comma 08 right parenthesis end cell row blank equals cell 11 plus log space 4 comma 08 end cell end table end style 

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Konsentrasi larutan barium asetat, Ba open parentheses C H subscript 3 C O O close parentheses subscript 2

table attributes columnalign right center left columnspacing 0px end attributes row M equals cell massa over M subscript r cross times 1000 over v end cell row blank equals cell fraction numerator 2 comma 55 over denominator 255 end fraction cross times 1000 over 500 end cell row blank equals cell 0 comma 02 end cell end table 

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Berdasarkan reaksi diatas, larutan garam ini bersifat basa, maka:

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 to the power of minus sign close square brackets 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 0 comma 02 cross times 2 end root end cell row blank equals cell square root of 4 cross times 10 to the power of negative sign 11 end exponent end root end cell row blank equals cell 2 cross times 10 to the power of negative sign 5 comma 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 blank equals cell negative sign log space 2 cross times 10 to the power of negative sign 5 comma 5 end exponent end cell row blank equals cell 5 comma 5 minus sign log space 2 end cell row cell pH and pOH end cell equals 14 row pH equals cell 14 minus sign left parenthesis 5 comma 5 minus sign log space 2 right parenthesis end cell row blank equals cell 8 comma 5 plus log space 2 end cell end table 

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Pembahasan Soal:

Kalium sianat (begin mathsize 14px style K O C N end style) merupakan garam basa yang mengalami hidrolisis parsial dengan reaksi sebagai berikut.


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begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell Mr space K O C N end cell equals cell 1 point Ar space K and 1 point Ar space O and 1 point Ar space C and 1 point Ar space N end cell row blank equals cell 1.39 plus 1.16 plus 1.12 plus 1.14 end cell row blank equals 81 row blank blank blank row cell M subscript anion end cell equals cell M subscript O C N to the power of minus sign end subscript double bond M subscript KOCN end cell row blank equals cell g over Mr cross times 1000 over mL end cell row blank equals cell fraction numerator 16 comma 2 over denominator 81 end fraction cross times 1000 over 200 end cell row blank equals 1 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 K subscript w over K subscript a cross times M subscript anion 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 6 end exponent cross times 1 end root end cell row blank equals cell square root of 10 to the power of negative sign 8 end exponent end root end cell row blank equals cell 10 to the power of negative sign 4 end exponent end cell end table end style


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Pembahasan Soal:

Larutan Na C N merupakan garam yang disusun oleh sisa basa kuat dan sisa asam lemah. Garam ini mengalami hidrolisis parsial dimana ion sisa asam lemah membentuk reaksi kesetimbangan dengan air sehingga melepas ion O H to the power of minus sign dengan reaksi sebagai berikut:

C N to the power of minus sign and H subscript 2 O equilibrium H C N and O H to the power of minus sign  

Berdasarkan reaksi diatas, larutan garam ini bersifat basa, maka:

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 to the power of minus sign 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 10 end exponent cross times 0 comma 01 end root end cell row blank equals cell square root of 10 to the power of negative sign 6 end exponent end root end cell row blank equals cell 10 to the power of negative sign 3 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 10 to the power of negative sign 3 end exponent end cell row blank equals 3 row cell pH and pOH end cell equals 14 row pH equals cell 14 minus sign 3 end cell row blank equals 11 end table  

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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

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