pH of $ 0.1\left( M \right)Na{H_2}P{O_4} $ is what?

Hint: pH is defined as the negative logarithm of the hydrogen ion concentration. As the given acid is a weak acid and has two hydrogen atoms. The concentration of hydrogen ions can be determined from the product of acid dissociation constant and given concentration. By substituting the obtained value in square root gives the concentration of hydrogen ions.
 $ pH = - \log \left( {{H^ + }} \right) $
 $ {H^ + } $ is hydrogen ion concentration.

Complete Step By Step Answer:
Given that concentration of $ Na{H_2}P{O_4} $ is $ 0.1M $
Sodium dihydrogen phosphate is a compound with the molecular formula of $ Na{H_2}P{O_4} $ , it is a weak acid consisting of two hydrogen atoms.
For strong acids, the hydrogen ion concentration will be the product of two and given concentration, i.e.., $ 2 \times 0.1M $
As the given acid is a weak acid, its concentration can be determined from the formula of $ \sqrt {{K_a} \times C} $ where $ {K_a} $ is the acid dissociation constant, and $ C $ is the given concentration.
The acid dissociation constant value is constant, which is $ 6.2 \times {10^{ - 8}} $
By substituting these two values, in the above formula,
 $ \sqrt {6.2 \times {{10}^{ - 8}} \times 0.1} $
Further simplifying the above value, it is equal to $ 7.9 \times {10^{ - 5}} $
The value obtained above is the final concentration of the hydrogen ion present in sodium dihydrogen phosphate.
Substitute this value in the pH formula,
 $ pH = - \log \left( {7.9 \times {{10}^{ - 5}}} \right) $
Further simplification, the value will be $ 4.10 $
Thus, the pH of $ 0.1\left( M \right)Na{H_2}P{O_4} $ is $ 4.10 $ .

Note:
In the above explanation, the concentration was determined from the value of acid dissociation constant and given concentration. By constructing an ICE table also, the final concentration of hydrogen ions can be determined. This value has to be substituted in the equation of pH.