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pH of a solution containing 0.25 M ammonia and 0.4M ammonium chloride solution given kb of NH3=1.8*10^-5

pH of a solution containing 0.25 M ammonia and 0.4M ammonium chloride solution given kb of NH3=1.8*10^-5

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Grade:12

1 Answers

Sahani Kumar
99 Points
5 years ago
Given: [salt(NH4Cl)]=0.44 [base(NH3)]=0.25. Kb=1.4*10^-5Now, pOH= – logKb+log[salt]/[Base] pOH= – log(1.4*10^-5)+log[0.44]/[0.25] pOH= (– log1.4)+ (– log10^-5)+(log1.76) pOH=( –0.146)+(5)+0.245 pOH=4.901Now we know that pH=pKw – pOH. where pKw=14 pH= 14 – 4.901 pOH=9.099

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