Question

what is a ph of 0.02N H2so4​

Answer 1

H2SO4(aq) ------> 2H+(aq) + (SO4)^2-(aq)

1 mole of H2SO4 produce 2 moles of H+

0.01 M of H2SO4 produce 0.02 M of H+

pH = -log[H+]

pH = -log(0.02) = 1.70

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Answer 2

The pH of the given solution is 2

Explanation:

We are given:

Normality of sulfuric acid = 0.02 N

The relation between normality and molarity follows:

$N=n\times M$

where,

N = normality of solution  = 0.02 N

n = n-factor of solution  = 2

M = molarity of solution = ?

Putting values in above equation, we get:

$0.02=2\times M\\\\M=\frac{0.02}{2}=0.01$

To calculate the pH of the solution, we use the equation:

$pH=-\log[H^+]$

We are given:

$[H^+]=0.01M$

Putting values in above equation, we get:

$pH=-\log(0.01)$

$pH=2$

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