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Pertanyaan

Suatu larutan penyangga dibuat melalui pencampuran 50,0 mL larutan natrium bikarbonat open parentheses Na H C O subscript 3 close parentheses 0,050 M dan 10,7 mL larutan NaOH 0,10 M. left parenthesis K subscript a space H C O subscript 3 to the power of minus sign equals 4 comma 69 cross times 10 to the power of negative sign 11 end exponent right parenthesis.

  1. Berapakah pH larutan penyangga?
  2. Berapa gram HCl yang harus ditambahkan ke dalam 25,0 mL larutan penyangga ini agar pH turun sebesar 0.07 satuan?space

Q. 'Ainillana

Master Teacher

Mahasiswa/Alumni Universitas Negeri Yogyakarta

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Pembahasan

Larutan penyangga yang dibuat dari 50,0 mL larutan natrium bikarbonat open parentheses Na H C O subscript 3 close parentheses 0,050 M dan 10,7 mL larutan NaOH 0,10 M merupakan larutan penyangga asam. NaOH akan bereaksi dengan H C O subscript 3 to the power of minus sign dari larutan Na H C O subscript 3

a. pH larutan penyangga:

Jumlah mol H C O subscript 3 to the power of minus sign yang bereaksi:

Na H C O subscript 3 left parenthesis italic a italic q right parenthesis yields Na to the power of plus sign left parenthesis italic a italic q right parenthesis plus H C O subscript 3 to the power of minus sign left parenthesis italic a italic q right parenthesis

Koefisien H C O subscript 3 to the power of minus sign = koefisien Na H C O subscript 3, maka:

n space H C O subscript 3 to the power of minus sign equals n space Na H C O subscript 3 n space H C O subscript 3 to the power of minus sign equals open square brackets Na H C O subscript 3 close square brackets middle dot V subscript Na H C O subscript 3 end subscript n space H C O subscript 3 to the power of minus sign equals 0 comma 050 space M middle dot 50 comma 0 space mL n space H C O subscript 3 to the power of minus sign equals 2 comma 5 space mmol

Jumlah mol 10,7 mL larutan NaOH 0,10 M:

open square brackets Na O H close square brackets equals fraction numerator n space Na O H over denominator V subscript NaOH end fraction n space Na O H double bond open square brackets Na O H close square brackets middle dot V subscript NaOH n space Na O H equals 0 comma 10 space M middle dot 10 comma 7 space mL n space Na O H equals 1 comma 07 space mmol

Konsentrasi H to the power of plus sign:

open square brackets H to the power of plus sign close square brackets double bond K subscript a fraction numerator n space H C O subscript 3 to the power of minus sign over denominator n space Na H C O subscript 3 end fraction open square brackets H to the power of plus sign close square brackets equals 4 comma 69 cross times 10 to the power of negative sign 11 end exponent fraction numerator 1 comma 43 space mmol over denominator 1 comma 07 space mmol end fraction open square brackets H to the power of plus sign close square brackets equals 6 comma 27 cross times 10 to the power of negative sign 11 end exponent

pH larutan:

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 6 comma 27 cross times 10 to the power of negative sign 11 end exponent pH equals 11 minus sign log space 6 comma 27 pH equals 10 comma 2

Jadi pH larutan penyangga yang terbentuk adalah 9,96.

b. pH larutan penyangga turun sebesar 0,07 satuan setelah penambahan HCl.

pH larutan setelah penambahan HCl:

pH equals 10 comma 2 minus sign 0 comma 07 pH equals 10 comma 13

Konsentrasi H to the power of plus sign dari larutan pH 9,89:

pH equals 10 comma 13 pH equals 11 minus sign 0 comma 87 pH equals 11 minus sign log space 10 to the power of 0 comma 87 end exponent pH equals 11 minus sign log space 7 comma 4 pH equals minus sign log space 7 comma 4 cross times 10 to the power of negative sign 11 end exponent pH equals minus sign log space open square brackets H to the power of plus sign close square brackets open square brackets H to the power of plus sign close square brackets equals 7 comma 4 cross times 10 to the power of negative sign 11 end exponent

Larutan HCl yang ditambahkan akan bereaksi dengan garam Na H C O subscript 3 pada larutan penyangga:

Jumlah mol HCl yang bereaksi adalah x mmol.

open square brackets H to the power of plus sign close square brackets double bond K subscript a fraction numerator n space H C O subscript 3 to the power of minus sign over denominator n space Na H C O subscript 3 end fraction 7 comma 4 middle dot 10 to the power of negative sign 11 end exponent equals 4 comma 69 middle dot 10 to the power of negative sign 11 end exponent fraction numerator left parenthesis 1 comma 43 plus x right parenthesis space mmol over denominator left parenthesis 1 comma 07 minus sign x right parenthesis space mmol end fraction 7 comma 4 left parenthesis 1 comma 07 minus sign x right parenthesis equals 4 comma 69 left parenthesis 1 comma 43 plus x right parenthesis 7 comma 918 minus sign 7 comma 4 x equals 6 comma 7076 plus 4 comma 96 x 12 comma 36 x equals 1 comma 2104 x equals fraction numerator 1 comma 2104 over denominator 12 comma 36 end fraction x equals 0 comma 1

Jadi jumlah mol HCl yang bereaksi adalah 0,1 mmol. Massa HCl (Mr = 36,5):

n space H Cl equals fraction numerator m space H Cl over denominator Mr space H Cl end fraction m space H Cl double bond n space H Cl middle dot Mr space H Cl m space H Cl equals 0 comma 1 space mmol middle dot 36 comma 5 m space H Cl equals 3 comma 65 space mg m space H Cl equals 3 comma 65 middle dot 10 to the power of negative sign 3 end exponent space g

Jadi massa HCl yang harus ditambahkan agar pH larutan turun sebesar 0,07 satuan adalah 3 comma 65 middle dot 10 to the power of negative sign 3 end exponent spacespace gram.

 

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