Match column I (effect of addition of 1M NaOH solution to 100 ml 1M HCl) with column II(P^H)

	column I		column II
A	50 ml of NaOH	I	log 3
B	90 ml of NaOH	II	14 - log 5
C	100 ml of NaOH	III	log 19
D	150 ml of NaOH	IV	7
		V	log 5

1.100 ml 1 M HCl +50 ml 1 M NaOH resulting solution is acidic

            $$\begin{gathered} \left[{\mathrm{H}}^{+}\right] =\frac{100\times 1-50\times 1}{100+50}=\frac{1}{3} \\ {\mathrm{p}}^{\mathrm{H} }=-\log \left(\frac{1}{3}\right)=\log 3 \end{gathered}$$

2. 100 ml 1 M HCl + 90 ml 1 M NaOH resulting solution is acidic

$$\begin{gathered} \left[{\mathrm{H}}^{+}\right] =\frac{100\times 1-90\times 1}{100+90}=\frac{1}{19} \\ {\mathrm{p}}^{\mathrm{H} }=-\log \left(\frac{1}{19}\right)=\log 19 \end{gathered}$$

3.100 ml 1 M HCl + 100 ml 1 M NaOH resulting solution is neutral ; p^H = 7

4.100 ml 1 M HCl + 150 ml 1 M NaOH resulting solution is basic

$$\begin{gathered} \left[{\mathrm{OH}}^{-}\right] =\frac{150\times 1-100\times 1}{100+150}=\frac{1}{5} \\ {\mathrm{p}}^{\mathrm{OH} }=-\log \left(\frac{1}{5}\right)=\log 5 \\ {\mathrm{p}}^{\mathrm{H}} = 14 - \log 5 \end{gathered}$$