Question

calculate pH of 0.005 molar H2 s o4​

Answer 1

Answer:

$\displaystyle{\text{pH}=2.31}$

Explanation:

We have given :

Concentration of Sulfuric acid :

$\displaystyle{\implies 0.005 \ \text{M \ or }5\times10^{-3}}$

We know :

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

Now putting concentration here :

$\displaystyle{\text{pH}=-\log{5\times10^{-3}}}\\\\\displaystyle{\text{pH}=-(\log{5+\log10^{-3}})}\\\\\displaystyle{\text{pH}=-(\log{5-3})}\\\\\displaystyle{\text{pH}=3-\log{5}}$

We know value of log 5 = 0.69

$\displaystyle{\text{pH}=3-0.69}\\\\\displaystyle{\text{pH}=2.31}$

Hence we get answer.