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Untuk mendapatkan larutan  () yang mempunyai pH = 10, tentukan banyaknya garam yang harus dilarutkan ke dalam 100 mL larutan (, )!

Pertanyaan

Untuk mendapatkan larutan begin mathsize 14px style C H subscript 3 C O O Na end style (begin mathsize 14px style M subscript r equals 82 end style) yang mempunyai pH = 10, tentukan banyaknya garam yang harus dilarutkan ke dalam 100 mL larutan (begin mathsize 14px style K subscript w equals 10 to the power of negative sign 14 end exponent end style, begin mathsize 14px style K subscript a space C H subscript 3 C O O H equals 10 to the power of negative sign 5 end exponent end style)!

Pembahasan Soal:

Perhatikan garam tersebut merupakan garam bersifat basa karena dibentuk dari asam lemah dan basa kuat, dimana sisa asam lemah akan bereaksi dengan air melepaskan ion undefined, sehingga:

1. Menghitung konsentrasi OH-

table attributes columnalign right center left columnspacing 0px end attributes row cell pH and pOH end cell equals 14 row pOH equals cell 14 minus sign 10 end cell row blank 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 open square brackets O H to the power of minus sign close square brackets end cell row cell space open square brackets O H to the power of minus sign close square brackets end cell equals cell 1 cross times 10 to the power of negative sign 4 end exponent end cell end table  

2. Menentukan konsentrasi anion garam melalui rumus hidrolisis garam

table attributes columnalign right center left columnspacing 0px end attributes row cell space 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 cell 1 cross times 10 to the power of negative sign 4 end exponent end cell equals cell square root of fraction numerator 1 cross times 10 to the power of negative sign 14 end exponent over denominator 1 cross times 10 to the power of negative sign 5 end exponent end fraction 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 cell open parentheses 1 cross times 10 to the power of negative sign 4 end exponent close parentheses squared end cell equals cell open parentheses square root of fraction numerator 1 cross times 10 to the power of negative sign 14 end exponent over denominator 1 cross times 10 to the power of negative sign 5 end exponent end fraction cross times open square brackets C H subscript 3 C O O to the power of minus sign close square brackets end root close parentheses squared end cell row cell 1 cross times 10 to the power of negative sign 8 end exponent end cell equals cell fraction numerator 1 cross times 10 to the power of negative sign 14 end exponent over denominator 1 cross times 10 to the power of negative sign 5 end exponent end fraction cross times open square brackets C H subscript 3 C O O to the power of minus sign close square brackets end cell row cell fraction numerator 1 cross times 10 to the power of negative sign 8 end exponent over denominator 1 cross times 10 to the power of negative sign 9 end exponent end fraction end cell equals cell open square brackets C H subscript 3 C O O to the power of minus sign close square brackets end cell row cell open square brackets C H subscript 3 C O O to the power of minus sign close square brackets end cell equals cell 10 space M end cell end table    

3. Menentukan konsentrasi garam

C H subscript 3 C O O Na yields C H subscript 3 C O O to the power of minus sign and Na to the power of plus sign

Karena koefisien begin mathsize 14px style C H subscript 3 C O O to the power of minus sign end style = undefined, maka konsentrasi garam undefined sama dengan konsentrasi anion garam begin mathsize 14px style C H subscript 3 C O O to the power of minus sign end style yaitu 10 M

4. Menentukan massa garam yang harus dilarutkan

begin mathsize 14px style 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 10 equals cell massa over 82 cross times 1000 over 100 end cell row massa equals cell 82 space gram end cell end table end style 

Jadi, massa garam tersebut adalah 82 gram.

Pembahasan terverifikasi oleh Roboguru

Terakhir diupdate 06 Mei 2021

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