Question

60ml of 0.1M NH4OH is mixed with 40ml of 0.1 M HCl. calculate pH of solution. pkb of NH4OH=4.74

Answer 1

I hope u understand. Thanks. for the question.

Answer 2

Answer: The pH of the solution is 8.96.

Explanation:-

$\text{no of moles}={\text{Molarity}\times {\text{Volume in L}}$

Thus $\text{no of moles}of NH_4OH={0.1M}\times {0.06 L}=0.006$

Thus $\text{no of moles}of HCl={0.1M}\times {0.04 L}=0.004$

$NH_4OH+HCl\rightarrow NH_4Cl+H_2O$

As 1 mole of $HCl$ combines with 1 mole of $NH_4OH$

0.004 moles of $HCl$ will combines with 0.004 mole of $NH_4OH$

Thus $HCl$ is the limiting reagent as it limits the formation of products and $NH_4OH$ is an excess reagent.

And 0.004 moles of $HCl$ will produce 0.004 moles of $NH_4Cl$

Moles of $NH_4OH$ left= (0.006-0.004) = 0.002 moles

The total volume of the solution will be

$V_{total}=60mL+40mL=100mL=0.1L$

According to the Henderson-Hasselbach equation, pH of such buffer solution can be calculated as:

$pOH=p_{K_b}+log\frac{[Salt]}{[base]}$

${[NH_4^+]}=\frac{moles}{\text {Volume in L}}=\frac{0.004}{0.1 L}=0.04M$

${[NH_4OH]}=\frac{moles}{\text {Volume in L}}=\frac{0.002}{0.1 L}=0.02M$

$pOH=4.74+log\frac{[0.04]}{[0.02]}$

$POH=5.04$

$pH=14-pOH$

$pH=14-(5.04)=8.96$