Question

$100ml$ of $1MHCl$ is mixed with $900ml$ of $01.MNaOH.$ $pH$ of the the final solution is:

• A. $\left[H^{+}\right]={10}^{-1}M$
• 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

Answer: (A)

$\left[H^{+}\right]={10}^{-1}M$

$V_a=100ml$ [acid]$=1M=1N\left[HCl\right]$

$V_b=900ml$ [base]$=0.1M=0.1N\left[NaOH\right]$

$Base⟶900\times 0.1=90Nml$

$Acid⟶100\times 1=100Nml$

So, remaining acid$=10Nml$ in $1000ml$ solution

$[Acid{]}_{solution}=\frac{10Nml}{1000ml}={10}^{-2}$

$⟹$ $pH=-log\left[H^{+}\right]=$ $-log{10}^{-1}=2$