Question

calculate the pH of sol. containing 0.98g H²SO⁴ in 100 mL of water​

Answer 1

answer is attached

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Answer 2

Answer:

0.6989

Explanation:

Given mass of H2SO4 = 0.98g

molecular mass of H2SO4 = 0.98g/mole

so, mole of H2SO4 = 0.98/98 = 0.01

Now concentration of H2SO4 = no. of mole × 1000/ volume of solution in ml

= 0.01 × 1000/100 = 0.1M

Dissociation of H2SO4 is .....

H2SO4 = 2H+ + SO4

So, concentration of H+ = 2 × concentration of H2SO4

= 0.2M

using formula, PH = -log[H+]

= -log0.2 = log(2×10^-1) = 1-log2

= 0.6989

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