KHP is a monoprotic acid which provides one H+ ion. How would your results be affected if a diprotic acid (such as sulfuric acid
, H2SO4) was used which provides two H+ ions? Calculate the molarity of the NaOH solution if 17.50 mL of N a O H was titrated against 10.50 mL of 0.75 M H2SO4.
In a diprotic acid, 2 moles of H+ ions is released. Therefore, number of moles of H+ in a diprotic acid = 2 × number of moles of H+ of monoprotic acid.
B.
Equation of the reaction
2NaOH + H2SO4 --> Na2SO4 + 2H2O
Number of moles of H2SO4 = molar concentration × volume
= 0.75 × 0.0105
= 0.007875 moles.
By stoichiometry, since 1 mole of H2SO4 reacts with 2 moles of NaOH. Therefore, number of moles of NaOH = 2 × 0.007875
= 0.01575 moles.
Molar concentration of NaOH = number of moles ÷ volume
Just imagine doing this experiment MULTIPLE TIMES and one of the trials you get 72 Kcal while in another u get 12kcal. It doesn't make sense. Somewhere in the experiment she went wrong. So its invalid
