The pH of milk, black coffee, tomato juice, lemon juice and egg white are 6.8, 5.0, 4.2, 2.2 and 7.8 respectively.

Solution:

The hydrogen ion concentration in the given substances can be calculated by using the given relation:

$\mathrm{pH}=-\log \left[\mathrm{H}^{+}\right]$

(i) $\mathrm{pH}$ of milk $=6.8$

Since, $\mathrm{pH}=-\log \left[\mathrm{H}^{+}\right]$

$6.8=-\log \left[\mathrm{H}^{+}\right]$

$\log \left[\mathrm{H}^{+}\right]=-6.8$

$\left[\mathrm{H}^{+}\right]=\operatorname{anitlog}(-6.8)$

$=1.5 \times 10^{-7} \mathrm{M}$

(ii) $\mathrm{pH}$ of black coffee $=5.0$

Since, $\mathrm{pH}=-\log \left[\mathrm{H}^{+}\right]$

$5.0=-\log \left[\mathrm{H}^{+}\right]$

$\log \left[\mathrm{H}^{+}\right]=-5.0$

$\left[\mathrm{H}^{+}\right]=\operatorname{anitlog}(-5.0)$

$=10^{-5} \mathrm{M}$

(iii) $\mathrm{pH}$ of tomato juice $=4.2$

Since, $\mathrm{pH}=-\log \left[\mathrm{H}^{+}\right]$

$4.2=-\log \left[\mathrm{H}^{+}\right]$

$\log \left[\mathrm{H}^{+}\right]=-4.2$

$\left[\mathrm{H}^{+}\right]=\operatorname{anitlog}(-4.2)$

$=6.31 \times 10^{-5} \mathrm{M}$

(iv) $\mathrm{pH}$ of lemon juice $=2.2$

Since, $\mathrm{pH}=-\log \left[\mathrm{H}^{+}\right]$

$2.2=-\log \left[\mathrm{H}^{+}\right]$

$\log \left[\mathrm{H}^{+}\right]=-2.2$

$\left[\mathrm{H}^{+}\right]=\operatorname{anitlog}(-2.2)$

$=6.31 \times 10^{-3} \mathrm{M}$

(v) $\mathrm{pH}$ of egg white $=7.8$

Since, $\mathrm{pH}=-\log \left[\mathrm{H}^{+}\right]$

$7.8=-\log \left[\mathrm{H}^{+}\right]$

$\log \left[\mathrm{H}^{+}\right]=-7.8$

$\left[\mathrm{H}^{+}\right]=\operatorname{anitlog}(-7.8)$

$=1.58 \times 10^{-8} \mathrm{M}$