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Titrasi larutan 25 ml HCl 0,1 M dengan 50 ml NaOH 0,1 M dengan indikator phenolftalein (PP).Hitunglah perubahan pH pada proses titrasi tersebut mulai dari penambahan 0 ml sampai dengan 50 ml .

Titrasi larutan 25 ml  0,1 M dengan 50 ml  0,1 M dengan indikator phenolftalein (PP). Hitunglah perubahan pH pada proses titrasi tersebut mulai dari penambahan 0 ml sampai dengan 50 ml Na O H

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perubahan pH titrasi pada penambahan0 ml sampai dengan 50 ml adalah seperti pada perhitungan di atas.

perubahan pH titrasi pada penambahan 0 ml sampai dengan 50 ml Na O H adalah seperti pada perhitungan di atas.

Pembahasan

pH larutan pada penambahan 0 ml : pH larutan pada penambahan 10 ml : pH larutan pada penambahan 20 ml : pH larutan pada penambahan 30 ml : pH larutan pada penambahan 40 ml : pH larutan pada penambahan 50 ml : Jadi, perubahan pH titrasi pada penambahan0 ml sampai dengan 50 ml adalah seperti pada perhitungan di atas.

pH larutan pada penambahan 0 ml Na O H:

H Cl yields H to the power of plus sign and Cl to the power of minus sign open square brackets H to the power of plus sign close square brackets double bond open square brackets H Cl close square brackets equals 0 comma 1 space M  pH equals minus sign log open square brackets H to the power of plus sign close square brackets space space space space equals minus sign log left parenthesis 0 comma 1 right parenthesis space space space space equals 1 

pH larutan pada penambahan 10 ml Na O H:

n space H Cl double bond M cross times V space space space space space space space space space equals 0 comma 1 space M cross times 25 space ml space space space space space space space space space equals 2 comma 5 space mmol n space Na O H double bond M cross times V space space space space space space space space space space space space space equals 0 comma 1 space M cross times 10 space ml space space space space space space space space space space space space space equals 1 space mmol 

 

open square brackets H Cl close square brackets equals fraction numerator n over denominator V space total end fraction space space space space space space space equals fraction numerator 1 comma 5 space mmol over denominator left parenthesis 25 plus 10 right parenthesis space ml end fraction space space space space space space space equals 0 comma 0429 space M open square brackets H to the power of plus sign close square brackets double bond open square brackets H Cl close square brackets equals 0 comma 0429 space M pH equals minus sign log open square brackets H to the power of plus sign close square brackets space space space space equals minus sign log space 0 comma 0429 space space space space equals 1 comma 37 

pH larutan pada penambahan 20 ml Na O H:

n space H Cl double bond M cross times V space space space space space space space space space equals 0 comma 1 space M cross times 25 space ml space space space space space space space space space equals 2 comma 5 space mmol n space Na O H double bond M cross times V space space space space space space space space space space space space space equals 0 comma 1 space M cross times 20 space mmol space space space space space space space space space space space space space equals 2 space mmol 

open square brackets H Cl close square brackets equals fraction numerator n over denominator V space total end fraction space space space space space space space space equals fraction numerator 0 comma 5 space mmol over denominator left parenthesis 25 plus 20 right parenthesis space ml end fraction space space space space space space space space equals 0 comma 0111 space M open square brackets H to the power of plus sign close square brackets double bond open square brackets H Cl close square brackets equals 0 comma 0111  pH equals minus sign log open square brackets H to the power of plus sign close square brackets space space space space equals minus sign log space 0 comma 0111 space space space space equals 1 comma 95

pH larutan pada penambahan 30 ml Na O H:

n space H Cl double bond M cross times V space space space space space space space space space equals 0 comma 1 space M cross times 25 space ml space space space space space space space space space equals 2 comma 5 space mmol n space Na O H double bond M cross times V space space space space space space space space space space space space equals 0 comma 1 space M cross times 30 space ml space space space space space space space space space space space space equals 3 space mmol 

 

open square brackets Na O H close square brackets equals fraction numerator n over denominator V space total end fraction space space space space space space space space space space space equals fraction numerator 0 comma 5 space mmol over denominator left parenthesis 25 plus 30 right parenthesis space ml end fraction space space space space space space space space space space space equals 0 comma 0091 space M open square brackets O H to the power of minus sign close square brackets double bond open square brackets Na O H close square brackets equals 0 comma 0091 space M  pOH equals minus sign log open square brackets O H to the power of minus sign close square brackets space space space space space space space equals minus sign log space 0 comma 0091 space space space space space space space equals 2 comma 04 pH equals 14 minus sign pOH space space space space equals 14 minus sign 2 comma 04 space space space space equals 11 comma 96 

pH larutan pada penambahan 40 ml Na O H:

n space H Cl double bond M cross times V space space space space space space space space space equals 0 comma 1 space M cross times 25 space ml space space space space space space space space space equals 2 comma 5 space mmol n space Na O H double bond M cross times V space space space space space space space space space space space space space equals 0 comma 1 space M cross times 40 space mmol space space space space space space space space space space space space space equals 4 space mmol 

 

open square brackets Na O H close square brackets equals fraction numerator n over denominator V space total end fraction space space space space space space space space space space space space equals fraction numerator 1 comma 5 space mmol over denominator left parenthesis 25 plus 40 right parenthesis space ml end fraction space space space space space space space space space space space space equals 0 comma 0231 space M open square brackets O H to the power of minus sign close square brackets double bond open square brackets Na O H close square brackets equals 0 comma 0231 space M  pOH equals minus sign log open square brackets O H to the power of minus sign close square brackets space space space space space space space equals minus sign log space 0 comma 0231 space space space space space space space equals 1 comma 636 pH equals 14 minus sign pH space space space space space equals 14 minus sign 1 comma 636 space space space space space equals 12 comma 36 

pH larutan pada penambahan 50 ml Na O H:

n space H Cl double bond M cross times V space space space space space space space space space equals 0 comma 1 space M cross times 25 space ml space space space space space space space space space equals 2 comma 5 space mmol n space Na O H double bond M cross times V space space space space space space space space space space space space space equals 0 comma 1 space M cross times 50 space ml space space space space space space space space space space space space space equals 5 space mmol 

open square brackets Na O H close square brackets equals fraction numerator n over denominator V space total end fraction space space space space space space space space space space space space equals fraction numerator 2 comma 5 space mmol over denominator left parenthesis 25 plus 50 right parenthesis space ml end fraction space space space space space space space space space space space space equals 0 comma 0333 space M open square brackets O H to the power of minus sign close square brackets double bond open square brackets Na O H close square brackets equals 0 comma 0333 space M  pOH equals minus sign log open square brackets O H to the power of minus sign close square brackets space space space space space space space equals minus sign log space 0 comma 0333 space space space space space space space equals 1 comma 477 pH equals 14 minus sign O H space space space space equals 14 minus sign 1 comma 477 space space space space equals 12 comma 52 

Jadi, perubahan pH titrasi pada penambahan 0 ml sampai dengan 50 ml Na O H adalah seperti pada perhitungan di atas.

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