Question

1.How many grams of NaOH required to be dissolved in a solution containing 0.05M H2SO4 so that the PH of will become 13
(molecular weight for NAOH is 40 g/mol)

2.When we add 0.1 mol from KOH to one liter of solution of 0.1M H2SO4,the PH of the resulting solution will be (1,2,4)?

Answer 1

1)

pH + pOH = 14

13 + pOH = 14

pOH = 1

pOH = -log(OH-) = 1

[OH-] = 0.1M

Since H2SO4 is a di-protic acid, so the reaction will be

2NaOH + H2SO4 ------ Na2SO4 + 2H2O

Assuming 1L of solution

Moles of NaOH required = 0.2 moles (Reason - 0.1 moles wil react with 0.05M H2SO4)

Mass of NaOH required = 0.2 mol * 40 gm/mol = 8 gms

2)

2KOH + H2SO4 -------- K2SO4 + 2H2O

Number of moles of KOH = 0.1

Number of moles of H2SO4 = Volume of solution (in L) * molarity = 1L * 0.1 M = 0.1 moles

moles of H2SO4 left = 0.1 - 0.1/2 = 0.05 moles

[H+] = 2 * numbe rof moles of H2SO4 = 2 * 0.05 = 0.1

pH = -log(H+) = -log(0.1) = 1