Question

Calculate pH when 9.8 g of H₂SO4 is dissolved in 2 L solution.




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Answer 1

Answer:

$\sf \: Molarity \: of \:  H_{2}SO_{4} = \frac{9.8 \times 1000}{98 \times 1000} = 0.1 \: mol$

As H2SO4 is strong acid so it dissociates completely to give H+ ions

$\sf \: H_{2}SO_{4} = 2 {H}^{ + } + SO_{4} ^{ 2- }$

[H+] = 2× 0.1 = 0.2 mol.

PH = -log[H+]

PH = 0.698