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Gas HCl murni, 18 mL dan gas NH3 murni 24 mL dilarutkan ke dalam 250 ml air sehingga seluruh gas larut dan tidak merubah volume air. Tekanan gas - gas semula 76 cmHg dan temperaturnya 27°C. Kalau tetapan (konstanta) gas ideal adalah R=0,08 L atm mol-1K-1, Kb NH4OH = 1 x 10-5, log 2 = 0,3; log 3 = 0,47 dan log 5 = 0,70 maka pH larutan tersebut adalah ...

A. Nurul

Master Teacher

Mahasiswa/Alumni UIN Syarif Hidayatullah Jakarta

Jawaban terverifikasi

Pembahasan

Volume HCl = 18 mL
Volume NH3 = 24 mL
P = 76 cmHg = 1 atm
R = 0.08
T = 27°C = 300K
menentukan mol masing masing :

begin mathsize 14px style nHCl equals fraction numerator P cross times V over denominator R cross times T end fraction nHCl equals fraction numerator 1 cross times 18 space mL over denominator 0 comma 08 cross times 300 K end fraction nHCl equals 18 over 24 nHCl equals 0 comma 75 space mmol end style 

Error converting from MathML to accessible text. 

menentukan reaksi MRS :

 

yang tersisa adalah basa lemah dengan asam konjugasinya, maka hitung dengan rumus buffer.

menentukan OH- :

size 14px open square brackets O H to the power of size 14px minus sign size 14px close square brackets size 14px space size 14px equals size 14px space size 14px Kb size 14px cross times fraction numerator size 14px mol size 14px space size 14px basa over denominator size 14px mol size 14px space size 14px garam end fraction size 14px open square brackets O H to the power of size 14px minus sign size 14px close square brackets size 14px space size 14px equals size 14px space size 14px 10 to the power of size 14px minus sign size 14px 5 end exponent size 14px cross times fraction numerator up diagonal strike size 14px 0 size 14px comma size 14px 25 end strike size 14px mmol over denominator up diagonal strike size 14px 0 size 14px comma size 14px 75 end strike size 14px mmol end fraction size 14px open square brackets O H to the power of size 14px minus sign size 14px close square brackets size 14px space size 14px equals size 14px 10 to the power of size 14px minus sign size 14px 5 end exponent size 14px cross times size 14px 1 over size 14px 3 size 14px open square brackets O H to the power of size 14px minus sign size 14px close square brackets size 14px space size 14px equals size 14px 10 to the power of size 14px minus sign size 14px 5 end exponent size 14px cross times size 14px 3 to the power of size 14px minus sign size 14px 1 end exponent 

menentukan pH :

begin mathsize 14px style 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 open square brackets 3 to the power of negative sign 1 end exponent cross times 10 to the power of negative sign 5 end exponent close square brackets pOH equals minus sign space left parenthesis log space left parenthesis 10 to the power of negative sign 5 end exponent right parenthesis space plus space log space left parenthesis 3 to the power of negative sign 1 end exponent right parenthesis right parenthesis pOH equals minus sign space left parenthesis minus sign 5 space minus sign space log space 3 right parenthesis pOH equals 5 space plus space log space 3 space pOH equals 5 plus 0 comma 47 pOH equals 5 comma 47  pH space space equals 14 minus sign pOH pH space space equals 14 minus sign 5 comma 47 pH space space equals 8 comma 53 end style

Jadi, pH larutan menjadi 8,53.

 

 

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