1) If it takes 50.0 mL of 0.50M KOH solution to completely neutralize 35mL of su
ID: 1032775 • Letter: 1
Question
1) If it takes 50.0 mL of 0.50M KOH solution to completely neutralize 35mL of sulfuric acid solution (H2SO4), what is the concentration of the H2SO4? (6 points) 2) If it takes 44.6mL of a 0.250M NaOH solution to completely neutralize 0.312g of an unknown diprotic acid, what is the molar mass of the acid? (6 points) 3) On the following titration curve, 14.0- 12.0 100F 8.0 6.0 4.0F 2.0 pH 5.0 10.0 15.0 20.0 25.0 30.0 Volume of 0.100 M NaOH, ml Label: (2 points each) The first equivalence point A The second equivalence point B What is the pH of A? What is the pH of B? Chemistry with VernierExplanation / Answer
1.
Balanced equation is,
2 KOH (aq.) + H2SO4 (aq.) ----------> K2SO4 (aq.) + 2 H2O (l)
Neutralisation formula,
M1 * V1 / n1 = M2 * V2 / n2
0.50 * 50.0 / 2 = M2 * 35 / 1
M2 = Concentration of Sulphuric acid = 0.357 M
2.
Balanced equation,
H2A (aq.) + 2 NaOH (aq.) ------------> Na2A (aq.) + 2 H2O (l)
Moles of NaOH = Molarity * Volume / 1000 = 0.250 * 44.6 / 1000 = 0.0112 mol
From the balanced equation,
2 mol of base needs 1 mol of acid
Then, 0.0112 mol of base needs 0.0112 * 1 / 2 = 0.00560 mol of diprotic acid
And
Number of moles = mass / molar mass
Molar mass of diprotic acid = 0.312 / 0.00560 = 55.7 g/mol
3.
pH of A = 6.0 (approx.)
pH of B = 12.0 (approx.)
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