Question

Which one of the following solutions will have pH close to unity?

• A. $100mL$ of $M/10HCl+100mL$ of $M/10NaOH$
• B. $55mL$ of $M/10HCl+45mL$ of $M/10NaOH$
• C. $10mL$ of $M/10HCl+90mL$ of $M/10NaOH$
• D. $75mL$ of $M/5HCl+25mL$ of $M/5NaOH$
• E. $50mL$ of $M/5HCl+50mL$ of $M/5NaOH$

Answer 1

Answer: (A), (C)

$100mL$ of $M/10HCl+100mL$ of $M/10NaOH$
$10mL$ of $M/10HCl+90mL$ of $M/10NaOH$

We need to calculate the resulting molar concentration of $H^{+}$ ions to determine the resulting pH of solutions.

Formula used is

$M_{mix}{V}_{mix}=M_{HCl}{V}_{HCl}+M_{NaOH}{V}_{NaOH}$

Option-A:

$V_{HCl}=100ml$, $M_{HCl}=0.1M$

$V_{NaOH}=100ml$,$M_{NaOH}=0.1M$

$M_{mix}{V}_{mix}=M_{HCl}{V}_{HCl}+M_{NaOH}{V}_{NaOH}$

$M_{mix}(100+100)=\frac{1}{10}\times 100+\frac{1}{10}\times 100=20$

$\Rightarrow M_{mix}=0.1M$

$pH=-log\left(H^{+}\right)=-log\left(0.1\right)=1$

Option-B:

$V_{HCl}=55ml$, $M_{HCl}=0.1M$

$V_{NaOH}=45ml$,$M_{NaOH}=0.1M$

$M_{mix}{V}_{mix}=M_{HCl}{V}_{HCl}+M_{NaOH}{V}_{NaOH}$

$M_{mix}(45+55)=\frac{1}{10}\times 100+\frac{1}{10}\times 100=20$

$\Rightarrow M_{mix}=0.2M$

$pH=-log\left(H^{+}\right)=-log\left(0.2\right)=0.698$

Option-C:

$V_{HCl}=10ml$, $M_{HCl}=0.1M$

$V_{NaOH}=90ml$,$M_{NaOH}=0.1M$

$M_{mix}{V}_{mix}=M_{HCl}{V}_{HCl}+M_{NaOH}{V}_{NaOH}$

$M_{mix}(10+90)=\frac{1}{10}\times 10+\frac{1}{10}\times 90=20$

$\Rightarrow M_{mix}=0.1M$

$pH=-log\left(H^{+}\right)=-log\left(0.1\right)=1$

Option-D:

$V_{HCl}=75ml$, $M_{HCl}=0.2M$

$V_{NaOH}=25ml$,$M_{NaOH}=0.2M$

$M_{mix}{V}_{mix}=M_{HCl}{V}_{HCl}+M_{NaOH}{V}_{NaOH}$

$M_{mix}(75+25)=\frac{2}{10}\times 75+\frac{2}{10}\times 25=20$

$\Rightarrow M_{mix}=0.2M$

$pH=-log\left(H^{+}\right)=-log\left(0.2\right)=0.698$

Option-E:

$V_{HCl}=50ml$, $M_{HCl}=0.2M$

$V_{NaOH}=50ml$,$M_{NaOH}=0.2M$

$M_{mix}{V}_{mix}=M_{HCl}{V}_{HCl}+M_{NaOH}{V}_{NaOH}$

$M_{mix}(50+50)=\frac{2}{10}\times 50+\frac{2}{10}\times 50=20$

$\Rightarrow M_{mix}=0.2M$

$pH=-log\left(H^{+}\right)=-log\left(0.2\right)=0.698$

Options A,C have the pH = 1.