Question

9. a If 25.0 mL of KOH solution is neutralized by 13.31 mL of 0.111 M H3PO4, what is the molarity of the KOH? State formula, of course you were going to do that, right? b) How many mL of 0.123 M H2SO4 solution is needed to neutralize 25,0 mL of 0.250 M KOH solution?

Answer 1

balanced reaction

H3PO4 + 3 KOH = 3 H2O + K3PO4

so 1 mole of H3PO4 is neutralized by 3 moles of KOH

moles of H3PO4 = molarity * volume in L = 0.111*0.01331 = 0.001477 moles

so moles of KOH required will be 3*0.001477 = 0.004432 moles

moalrity is number of moles in 1L of solution

so now we have 25 mL of KOH solution in which 0.004432 moles are there

hence in 1L or 1000 mL moles will be (0.004432 /25) * 1000 = 0.1772 M

(b) H2SO4 + 2KOH = K2SO4 + 2H2O

so 1 mole of H2SO4 is neutralized by 2 moles of KOH

moles of H2SO4 = molarity * volume in L = 0.25*0.025= 0.00625 moles

so moles of KOH required will be 2*0.00625= 0.0125moles

moles = molarity * volume in L

0.0125 = 0.123 * volume

volume = 0.1016 L or 101.62 mL