Question

Calculate pH of a solution which contains 9.9 mL of 1 M $HCl$ and 100 mL of 0.1 M $NaOH$.
Take $log\left(9.1\right)=0.96$

• A. 7.96
• B. 12.96
• C. 10.96
• D. 5.96

Answer 1

The correct option is C 10.96

The reaction proceeds as :
$NaOH+HCl\rightleftharpoons NaCl+H_{2}O$
As we know,
1 mole of $HCl$ reacts with the 1 mole of $NaOH$.
Lets find the moles of solution,
Moles of $HCl$ in 9.9 ml of 1 M solution = $\frac{9.9}{1000}\times 1=0.0099molHCl$
Moles of NaOH in 100 ml of 0.1 M solution =$\frac{100}{1000}\times 0.1=0.01molNaOH$
On mixing the 0.0099 mole of $HCl$ will be neutralised = 0.01 - 0.0099 =$1\times {10}^{-4}$ mol $NaOH$ dissolved in 109.9 mL solution.
Hence, the molarity of $NaOH$ solution=$[OH{]}^{-}=\frac{1\times {10}^{-4}mol}{\frac{109.9}{1000}L}=9.099\times {10}^{-4}MpOH=-log[OH{]}^{-}=-log(9.099\times {10}^{-4})pOH=-(0.96-4)=3.04pH+pOH=14pH=14-3.04=10.96$