Question

Determine the $pH$ of the solution that results from the addition of $20.00mL$ of $0.01MCa(OH{)}_2$ to $30.00mL$ of $0.01MHCl$:

• A. $11.30$
• B. $10.53$
• C. $2.70$
• D. $8.35$

Answer 1

The correct option is C $2.70$

$Ca(OH{)}_2+HCl=CaC{l}_2+2H_{2}O$

$HCl$ is strong acid, so number of moles of $HCl$= Number of moles of $H^{+}$

In $30ml$ of $0.01M$ $HCl$; moles= $L.M$

$=30\times {10}^{-3}\times 0.01$

$=0.0003$ moles= $H^{+}$

In $20ml$ of $0.01M$ $Ca(OH{)}_2$= $20\times {10}^{-3}\times 0.01=0.0002$ moles= $O{H}^{-}$

Now remaining mole of $H^{+}=0.0003-0.0002=0.0001$ of $H^{+}$

Total solution= $50ml(30+20=50)$

Concentration= $\frac{0.0001}{0.05L}=0.002M$ of $H^{+}$

$\Rightarrow p^H=-\log \left[H^{+}\right]\Rightarrow -\log \left[0.002\right]$

$\Rightarrow p^H=2.698\approx 2.7$