The equation for the reaction of oxalic acid with a solution of caustic potash. 2‌KOH(aq)+H2C2O4⋅2H2O(aq) ⟶K2C2O4(aq)+4H2O Molarity equation, ‌
M1V1
n1
=‌
M2V2
n2
where M1= molarity of acid =5 millimol M2= molarity of base =5 millimol V1= volume of acid =100‌mL ‌V2=‌ volume of base ‌=100‌mL ‌n1=‌ stoichiometric coefficient of acid ‌=1 ‌n2=‌ stoichiometric coefficient of base ‌=2 As it is clear from balanced equation that 2 moles of KOH are required to neutralize one mole of oxalic acid. Thus, the solution formed by 5 millimoles of KOH and 5 millimoles of oxalic acid mixed and dissolved in 100‌mL water will be acidic.