Answer to Question #192132 in General Chemistry for Yasemin

Question

What is the pH when 20 ml 0.1 M HCl is added to 100 ml 0.1M NaCN?

Answer 1

$HCl+NaCN \Rightarrow HCN+ NaCl$

Now initially the moles of each:

$n_{HCl}=0.02\times0.1=0.002 moles$

$n_{NaCN}=0.1\times0.1=0.01moles$

from above we can see that HCl is a limiting reagent,so it consumes completely.

Hence after the reaction

$n_{HCl}=0$

$n_{NaCN}=$ $0.01-0.002=0.008moles$

SO concentration of $NaCN=\dfrac{n_{NaCN}}{Total volume}=\dfrac{0.008}{0.02+0.1}=0.067M$

As $NaCN$ is a base so it will gave us $P_{_{OH}}$

Hence $P_{OH}=-log[CN^-]$

$P_{OH}=-log[0.067]=1.17M$

So $P_{H}=14-1.17=12.83M$