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50 mL larutan CH3​COOH 0,5 M (Ka​=10−5) direaksikan dengan 50 mL larutan NaOH 0,5 M. Besarnya pH campuran ialah ...

Pertanyaan

50 mL larutan C H subscript 3 C O O H 0,5 M (K subscript a equals 10 to the power of negative sign 5 end exponent) direaksikan dengan 50 mL larutan NaOH 0,5 M. Besarnya pH campuran ialah ...

N. Puspita

Master Teacher

Jawaban terverifikasi

Jawaban

jawaban yang benar pH campuran adalah 9 + log 1,58.

Pembahasan

Menentukan nilai mol larutan:

table attributes columnalign right center left columnspacing 0px end attributes row cell n subscript C H subscript 3 C O O H end subscript end cell equals cell M cross times V end cell row blank equals cell 0 comma 5 cross times 50 end cell row blank equals cell 25 space mmol end cell row blank equals cell 0 comma 025 space mol end cell end table 

table attributes columnalign right center left columnspacing 0px end attributes row cell n subscript NaOH end cell equals cell M cross times V end cell row blank equals cell 0 comma 5 cross times 50 end cell row blank equals cell 25 space mmol end cell row blank equals cell 0 comma 025 space mol end cell end table 

Persamaan reaksi:

 

Menentukan nilai pH campuran:

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 0 comma 025 space mol space space space space space space space 0 comma 025 space mol

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 K subscript w over K subscript a cross times fraction numerator mol space C H subscript 3 C O O to the power of minus sign over denominator left parenthesis V space C H subscript 3 C O O H and V space Na O H right parenthesis end fraction 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 fraction numerator 0 comma 025 space mol over denominator 0 comma 1 space L end fraction 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 25 end root end cell row blank equals cell square root of 2 comma 5 cross times 10 to the power of negative sign 10 end exponent end root end cell row blank equals cell 1 comma 58 cross times 10 to the power of negative sign 5 end exponent space M end cell end table 

table attributes columnalign right center left columnspacing 0px end attributes row cell italic p O H end cell 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 58 cross times 10 to the power of negative sign 5 end exponent end cell row blank equals cell 5 minus sign log space 1 comma 58 end cell end table 

table attributes columnalign right center left columnspacing 0px end attributes row cell italic p H end cell equals cell 14 minus sign italic p O H end cell row blank equals cell 14 minus sign left parenthesis 5 minus sign log space 1 comma 58 right parenthesis end cell row blank equals cell 9 plus log space 1 comma 58 end cell end table

Jadi, jawaban yang benar pH campuran adalah 9 + log 1,58.

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Seorang siswa membuat campuran 100 ml asam sianida (HCN) 0,1 M dan 200 ml kalium hidroksida (KOH) 0,05 M. (Ka​ HCN=5×10−10) a. Tuliskan reaksi yang terjadi ! b. Hitunglah pH larutan tersebut !

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Jawaban terverifikasi

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