Direaksikan 100 ml larutan KOH 0,4 M dengan 100 ml larutan CH3COOH 0,4 M (KaCH3COOH=1,8×10−5). Tentukan pH campuran tersebut !

Pertanyaan

Direaksikan 100 ml larutan $\style{font-size:14px}{\mathrm{KOH}}$ 0,4 M dengan 100 ml larutan $\style{font-size:14px}{{\mathrm{CH}}_3\mathrm{COOH}}$ 0,4 M $\style{font-size:14px}{({\mathrm K}_{\mathrm a}\;{\mathrm{CH}}_3\mathrm{COOH}=1,8\times10^{-5})}$ . Tentukan pH campuran tersebut ! $\;$

Pembahasan Soal:

Campuran di atas merupakan campuran asam lemah dengan basa kuat. Untuk mengetahui apakah campuran tersebut merupakan larutan penyangga atau garam yang terhidrolisis maka dapat direaksikan sebagai berikut:

$begin mathsize 14px style mol space K O H double bond V space K O H space cross times M space K O H mol space K O H equals 100 space mL cross times 0 comma 4 space M mol space K O H equals 40 space mmol mol space C H subscript 3 C O O H double bond V space C H subscript 3 C O O H cross times M space C H subscript 3 C O O H mol space C H subscript 3 C O O H equals 100 space mL cross times 0 comma 4 space M mol space C H subscript 3 C O O H equals 40 space mmol end style$

Pada reaksi tersebut asam lemah dan basa kuat habis bereaksi dan yang tersisa adalah garamnya dan air. Garam tersebut dalam air akan akan terionisasi sempurna menghasilkan kation dan anion. Anion dari asam lemah dan kation dari basa kuat.

$begin mathsize 14px style C H subscript 3 C O O K left parenthesis italic a italic q right parenthesis yields with plus air on top C H subscript 3 C O O to the power of minus sign left parenthesis italic a italic q right parenthesis plus K to the power of plus sign left parenthesis italic a italic q right parenthesis end style$

$begin mathsize 14px style C H subscript 3 C O O to the power of minus sign end style$ akan bereaksi dengan air (terhidrolisis) sesuai dengan persamaan berikut.

$begin mathsize 14px style C H subscript 3 C O O to the power of minus sign left parenthesis italic a italic q right parenthesis plus H subscript 2 O open parentheses italic l close parentheses equilibrium C H subscript 3 C O O H left parenthesis italic a italic q right parenthesis plus O H to the power of minus sign left parenthesis italic a italic q right parenthesis end style$

Adanya ion $begin mathsize 14px style O H to the power of minus sign end style$ dalam hasil reaksi menunjukan bahwa larutan garam tersebut bersifat basa maka pH dapat dicari dengan persamaan berikut:

$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 italic h cross times open square brackets garam 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 K subscript italic w over K subscript italic a cross times open square brackets fraction numerator mol space C H subscript 3 C O O K over denominator V space larutan end fraction 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 fraction numerator 10 to the power of negative sign 14 end exponent over denominator 1 comma 8 cross times 10 to the power of negative sign 5 end exponent end fraction cross times open square brackets fraction numerator 40 space mmol over denominator 200 space mL end fraction 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 fraction numerator 10 to the power of negative sign 14 end exponent over denominator 1 comma 8 cross times 10 to the power of negative sign 5 end exponent end fraction cross times 2 cross times 10 to the power of negative sign 1 end exponent space M end root end cell row cell open square brackets O H to the power of minus sign close square brackets end cell equals cell 1 comma 05 cross times 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 1 comma 05 cross times 10 to the power of negative sign 5 end exponent space M close square brackets end cell row pOH equals cell 5 minus sign log space 1 comma 05 end cell row blank blank blank row pH equals cell 14 minus sign pOH end cell row pH equals cell 14 minus sign left parenthesis 5 minus sign log space 1 comma 05 right parenthesis end cell row pH equals cell 9 plus log space 1 comma 05 end cell end table end style$

Jadi pH campuran tersebut adalah $begin mathsize 14px style bold 9 bold plus bold log bold space bold 1 bold comma bold 05 end style$ . $space$

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

L. Avicenna

Mahasiswa/Alumni Institut Teknologi Bandung

Terakhir diupdate 06 Oktober 2021