stoichiometry...

Question

KHC8H4O4(aq) + NaOH(aq) → H2O(l) + KNaC8H4O4(aq)

a) How many moles of NaOH (aq) is required to neutralize 0.8000 g of ‘KHP’ in the titration?

b) It took 20.00 mL of NaOH (aq) to neutralize ’KHP’. Calculate the molarity of the NaOH(aq) solution.

H2A(aq) + 2NaOH(aq) → 2H2O(l) + Na2A(aq)

a) 22.20 mL of NaOH (aq) was needed to neutralize acid (hint: molarity of NaOH (aq) is the value that you calculated in question 2b). How many moles of NaOH (aq) was used?

b) How many moles of acid, H2A, was neutralized?

c) 0.2500 g of acid was weighed out for neutralization. Calculate the molar mass (g/mol) of the acid

I do not know how to do stoichiometry yet but I have this question due soon. If I am missing anything you need let me know please.

Answer 1

First question,

(a) assuming NaOH reacts to KHP in a 1:1 ratio,

(0.8g KHP) / (204.22g/mol KHP) * (1 mol NaOH / 1 mol KHP) = 3.917 x 10^-3 mol NaOH

(b) (3.917 x 10^-3 mol NaOH) / (20 mL) * (1000mL/1 L) = 1.959 x 10^-1 M

Second question

(a) 3.917 x 10^-3 mol NaOH, assuming molarity is same as (b) above

(b) 1.959 x 10^-3 mole H₂A

(c) 0.2500 g / 1.959 x 10^-3 mole H₂A = 1.276 x 10² g/mol H₂A