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Tentukan pH larutan sebelum dicampur dan sesudah dicampur jika 100 mL larutan  0,1 M direaksikan dengan 100 mL larutan  0,1 M!

Pertanyaan

Tentukan pH larutan sebelum dicampur dan sesudah dicampur jika 100 mL larutan H subscript bold 2 S O subscript bold 4 0,1 M direaksikan dengan 100 mL larutan K O H 0,1 M!space 

Pembahasan Soal:

Menentukan Nilai pH sebelum dicampur

pH H subscript bold 2 S O subscript bold 4

H subscript 2 S O subscript 4 adalah asam kuat yang terionisasi sempurna di dalam air menurut reaksi berikut:

H subscript 2 S O subscript 4 left parenthesis italic a italic q right parenthesis yields 2 H to the power of plus sign left parenthesis italic a italic q right parenthesis plus S O subscript 4 to the power of 2 minus sign left parenthesis italic a italic q right parenthesis 

Nilai konsentrasi ion H to the power of plus sign dapat ditentukan secara hitungan stoikiometri sesuai jumlah koefisien ion ion H to the power of plus sign .

table attributes columnalign right center left columnspacing 0px end attributes row cell open square brackets H to the power of plus sign close square brackets end cell equals cell a cross times M subscript a end cell row cell open square brackets H to the power of plus sign close square brackets end cell equals cell 2 cross times 10 to the power of negative sign 1 end exponent space M end cell row blank blank blank row pH equals cell negative sign log space open square brackets H to the power of plus sign close square brackets end cell row pH equals cell negative sign log space open square brackets 2 cross times 10 to the power of negative sign 1 end exponent space M close square brackets end cell row pH equals cell 1 minus sign log space 2 end cell end table 

Jadi pH H subscript bold 2 S O subscript bold 4 sebelum dicampur adalah bold 1 bold minus sign bold log bold space bold 2.
 

pH KOH

 K O H adalah basa kuat yang terionisasi sempurna di dalam air menurut reaksi berikut:

K O H left parenthesis italic a italic q right parenthesis yields K to the power of plus sign 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  

Nilai konsentrasi ion O H to the power of minus sign dapat ditentukan secara hitungan stoikiometri sesuai jumlah koefisien ion O H to the power of minus sign .

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 b cross times M subscript italic b end cell row cell 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 1 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 cross times 10 to the power of negative sign 1 end exponent space M close square brackets end cell row pOH equals 1 row blank blank blank row pH equals cell 14 minus sign pOH end cell row pH equals cell 14 minus sign 1 end cell row pH equals 13 end table 
Jadi
 pH K O H sebelum dicampur adalah 13.

Menentukan pH campuran

table attributes columnalign right center left columnspacing 0px end attributes row cell mol space H subscript 2 S O subscript 4 end cell equals cell V space H subscript 2 S O subscript 4 cross times M space H subscript 2 S O subscript 4 end cell row cell mol space H subscript 2 S O subscript 4 end cell equals cell 100 space mL cross times 0 comma 1 space M end cell row cell mol space H subscript 2 S O subscript 4 end cell equals cell 10 space mmol end cell row blank blank blank row cell mol space K O H end cell equals cell V space K O H cross times M space K O H end cell row cell mol space K O H end cell equals cell 100 space mL cross times 0 comma 1 space mol end cell row cell mol space K O H end cell equals cell 10 space mmol end cell end table 
 


 

Larutan yang tersisa adalah H subscript 2 S O subscript 4, maka larutan campuran akan bersifat asam dan pH-nya dapat dicari dengan penyelesaian berikut:

open square brackets H to the power of plus sign close square brackets equals a cross times M subscript a open square brackets H to the power of plus sign close square brackets equals a cross times mol subscript sisa over V subscript total open square brackets H to the power of plus sign close square brackets equals 2 cross times fraction numerator 5 cross times 10 to the power of negative sign 3 end exponent space mol over denominator 2 cross times 10 to the power of negative sign 1 end exponent space L end fraction open square brackets H to the power of plus sign close square brackets equals 5 cross times 10 to the power of negative sign 2 end exponent space M  pH equals minus sign log space open square brackets H to the power of plus sign close square brackets pH equals minus sign log space open square brackets 5 cross times 10 to the power of negative sign 2 end exponent space M close square brackets pH equals 2 minus sign log space 5  

Jadi pH larutan setelah dicampur adalah bold 2 bold minus sign bold log bold space bold 5.space 

Pembahasan terverifikasi oleh Roboguru

Terakhir diupdate 04 Mei 2021

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