Question

Calculate the ph of a solution containing 0.98 g of h2so4 in 100 ml of water

Answer 1

given, mass of H2SO4 = 0.98g

molecular mass of H2SO4 = 98g/mol

so, mole of H2SO4 = 0.98/98 = 0.01

now, concentration of H2SO4 = number of mole × 1000 /volume of solution in ml

= 0.01 × 1000/100 = 0.1 M

dissociation of H2SO4 is .....

$H_2SO_4\Leftrightarrow2H^++SO_4^{--}$

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

hence, pH of H2SO4 is 0.6989