Pertanyaan

Larutan Ba(HCOO) 2 0,04 M, Ka HCOOH = 8 × 1 0 − 5 terhidrolisis sebagian/parsial, jikaMr Ba(HCOO) 2 = 228. Tentukan : a. pH larutan b. derajad hidrolisis c. massaBa(HCOO) 2 yang terlarut dalam 500 ml larutan

Larutan Ba(HCOO)₂ 0,04 M, Ka HCOOH = $8 \times 1 0^{- 5}$ terhidrolisis sebagian/parsial, jika Mr Ba(HCOO)₂= 228. Tentukan :

a. pH larutan

b. derajad hidrolisis

c. massa Ba(HCOO)₂ yang terlarut dalam 500 ml larutan

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A. Nurul

Master Teacher

Mahasiswa/Alumni UIN Syarif Hidayatullah Jakarta

Jawaban terverifikasi

Jawaban

pH larutan = 8+log 2,33,derajad hidrolisis = , danmassaBa(HCOO) 2 = 5,76 gram.

pH larutan = 8+log 2,33, derajad hidrolisis =

, dan massa Ba(HCOO)₂ = 5,76 gram.

Pembahasan

a. menentukan pH garam basa : b. menentukan derajat hidrolisis : c. menentukan massaBa(HCOO) 2 yang terlarut dalam 500 ml larutan: Jadi, pH larutan = 8+log 2,33,derajad hidrolisis = , danmassaBa(HCOO) 2 = 5,76 gram.

a. menentukan pH garam basa :

$begin mathsize 14px style open square brackets O H to the power of minus sign close square brackets equals square root of Kw over Ka cross times open square brackets garam close square brackets end root open square brackets O H to the power of minus sign close square brackets equals square root of fraction numerator 10 to the power of negative sign 14 end exponent over denominator 8 cross times 10 to the power of negative sign 5 end exponent end fraction cross times open square brackets 0 comma 04 close square brackets end root open square brackets O H to the power of minus sign close square brackets equals square root of fraction numerator 10 to the power of negative sign 14 end exponent over denominator 8 cross times 10 to the power of negative sign 5 end exponent end fraction cross times open square brackets 4 cross times 10 to the power of negative sign 2 end exponent close square brackets end root open square brackets O H to the power of minus sign close square brackets equals square root of fraction numerator 4 cross times 10 to the power of negative sign up diagonal strike 16 end exponent over denominator 8 cross times 10 to the power of up diagonal strike negative sign 5 end strike end exponent end fraction end root open square brackets O H to the power of minus sign close square brackets equals square root of 5 cross times 10 to the power of negative sign 12 end exponent end root open square brackets O H to the power of minus sign close square brackets equals 2 comma 23 space cross times 10 to the power of negative sign 6 end exponent pOH equals minus sign log space open square brackets O H to the power of minus sign close square brackets pOH equals minus sign log space left square bracket 2 comma 23 space cross times 10 to the power of negative sign 6 end exponent right square bracket pOH equals 6 minus sign log space 2 comma 23 pH space space equals 14 minus sign pOH pH space space equals 14 minus sign left parenthesis 6 minus sign log space 2 comma 23 right parenthesis pH space space equals 8 plus log space 2 comma 23 end style$

b. menentukan derajat hidrolisis :

$begin mathsize 14px style Kh space equals Kw over Ka Kh space equals fraction numerator 10 to the power of negative sign 14 end exponent over denominator 8 cross times 10 to the power of negative sign 5 end exponent end fraction Kh space equals 0 comma 125 cross times 10 to the power of negative sign 9 end exponent Kh space equals 125 space cross times space 10 to the power of negative sign 12 end exponent end style$

$begin mathsize 14px style proportional to equals fraction numerator Kh over denominator open square brackets garam close square brackets end fraction proportional to equals fraction numerator 125 cross times 10 to the power of negative sign 12 end exponent over denominator left square bracket 4 cross times 10 to the power of negative sign 2 end exponent right square bracket end fraction proportional to equals fraction numerator 125 cross times 10 to the power of negative sign 12 end exponent over denominator left square bracket 4 cross times 10 to the power of negative sign 2 end exponent right square bracket end fraction proportional to equals 31 comma 25 cross times 10 to the power of negative sign 10 end exponent end style$

c. menentukan massa Ba(HCOO)₂ yang terlarut dalam 500 ml larutan:

$begin mathsize 14px style M space space space space space space space space space equals massa over Mr cross times 1000 over V 4 cross times 10 to the power of negative sign 2 end exponent equals massa over 288 cross times fraction numerator up diagonal strike 1000 squared over denominator up diagonal strike 500 end fraction massa space space space equals fraction numerator up diagonal strike 4 squared cross times 10 to the power of negative sign 2 end exponent cross times 288 over denominator up diagonal strike 2 end fraction massa space space equals 5 comma 76 space gram end style$

Jadi, pH larutan = 8+log 2,33, derajad hidrolisis = , dan massa Ba(HCOO)₂ = 5,76 gram.

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