Question

If 25.0 mL of 0.150 M NaOH were used to titrate a solution of H2SO4, how many grams of H2SO4 are present in the solution?

Answer 1

Given molarity of NaOH = 0.15M so, Normality of NaOH= molarity x Valancy factor - 0:15 x 1 N = 0.15 N a s valency factor 2 for Ngo H=17 Volume = 25.0 ml = 0.025L We know that at the end of titration equivalent of acid = equivalent of base so, equivalent of aciter = normality of base x volume (in 1) Ou5 N X 0.025L 0.00375 equivalent -

we know that it number of equivalent = molex valency - 0'00375 equivalent a boole x 2 so, mol of H2SO4= 0.00375 mol factor (2 valency factor 2 for H2s04=2 ] o nbevod label - 0.001875 mol s : mass of H₂SO4= mol x Mo -0.001875 mol x 98 g/mol = 01183759