Question

2) If a student weighs out 0.500 g o tudent weighs out 0.500 g of KHP and titrates it with sodium hydroxide solution, the molarity of the sodium hydroxide solution if it takes 31.21 mL of it to titrate the KHP? It takes 26.61 mL of 0.228 M sodium hydroxide solution to titrate 10.00 mL of sulfuric acid solution. What is the molarity of the sulfuric acid?

Answer 1

2-

The given KHP (Potassium hydrogen phthalate) is an acid salt compound of the weak acid phthalic acid (C₈H₆O₄)

Now if we titrate KHP with a strong base like NaOH, that means we are titrating the weak acid phthalic acid with a strong base.

i.e KHP + NaOH ------------> salt + H2O

Again KHP dissociate in water as - KHP -------------> K+ + HP- or K+ + C₈H₅O₄^-

Thus we can write

C₈H₅O₄^- + NaOH -----------------> C₈H₅O₄Na (salt)

This indicates that to neutralize 1 mole of KHP, we need 1 mole of NaOH

Now given mass of KHP taken = 0.500 g

We know molar mass of KHP = 204.22 g/mol

Thus moles of KHP taken = mass / molar mass

= 0.500 g / 204.22 g/mol

= 0.0024 moles

Thus moles of NaOH required to neutralize this = 0.0024 moles

Given volume of KOH taken = 31.21 ml = 0.03121 L

Thus concentration of KHP taken = moles / volume in L

= 0.0024 moles / 0.03121 L

= 0.077 M

2-

The reaction between Sulphuric acid (H2SO4) and NaOH is

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

That means we need 2 moles of NaOH to neutralize 1 mole of H2SO4

Now given NaOH taken = 26.61 ml and 0.288 M

Thus moles of NaOH taken = concentration * volume

= 0.288 M * 26.61 ml

= 0.288 mol/ 1000 ml * 26.61 ml

= 0.0076 moles

Since 2 moles of NaOH neutralize 1 mole of H2SO4

Then 0.0076 moles of NaOH neutralize = 1/2 * 0.0076

= 0.0038‬ mole of H2SO4

Now given volume of H2SO4 taken = 10 ml = 0.010 L

Thus concentration ofH2SO4 taken = moles / volume in L

= 0.0038 moles /0.010 L

= 0.38 M