Question

100 ml of 1M $HCl$ is mixed with 900 ml of 0.1 M $NaOH$. In the final solution:

• A. $\left[H^{+}\right]={10}^{-2}M$
• B. $\left[C{l}^{-}\right]={10}^{-1}M$
• C. $\left[N{a}^{+}\right]=\left[C{l}^{-}\right]$
• D. $\left[O{H}^{-}\right]={10}^{-2}M$

Answer 1

The correct option is A $\left[H^{+}\right]={10}^{-2}M$

$NaOH+HCl\to NaCl+H_{2}O$

$1$ mole $1$ mole

Moles of $HCl$ in $100ml$ of $1M$ $HCl$

$\Rightarrow \frac{100\times 1}{100}=0.1$ mole of $HCl$

Mole of $NaOH$ in $900ml$ of $0.1M$ $NaOH$

$\Rightarrow \frac{900}{1000}\times 0.1=0.09$ mole $NaOH$

On mixing $0.09$ mole of $NaOH$ will be neutralised

$=0.1-0.09$

$=0.01$ mole of $HCl$ in $1000ml$ solution

Hence molarity of $HCl=\frac{0.01}{1000}\times 1000=0.01M$

$\therefore \left[H^{+}\right]=0.01={10}^{-2}M$