# Acid – Base Review Problems

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Acid – Base Review Problems
1. H2SO4 + 2 KOH → K2SO4+ 2 H2O 2 HCl + Ca(OH)2 → CaCl2 + 2 H2O 2 H3PO4 + 3 Sr(OH)2 → 6 H2O + Sr3(PO4)2 d) 2 HNO3 + Mg(OH)2 → 2 H2O + Mg(NO3)2

2. Determine the pH and pOH of
A) M KOH [OH-] = 1 x 10-3 pOH = pH = 11.0 B) M HCl [H+] = 1 x 10-4 pH = pOH = 10.0

3. pH = -log [H+] pH + pOH = 14 pOH = -log [OH-] [H+] = 5.0 x 10-6 pH = 5.3 b) [OH-] = 2.0 x 10-5 pOH = 4.7 pH = 9.3 c) [H+] = 8.3 x pH = 9.1 d) [OH-] = 4.5 x pOH = 10.3 pH = 3.7

4. -log [10-X] = X a) pH = 4 [H+] = 1 x 10-4 M b) pH = 11 [H+] = 1 x M c) pH = 8 [H+] = 1 x 10-8 M 5. pH + pOH = 14 a) pH=6 pOH = 8 [OH-] = 1 x 10-8 M b) pH =9 pOH = 5 [OH-] = 1 x 10-5 M c) pH = 12 pOH = 2 [OH-] = 1 x 10-2 M

6. Calculate the molarity of 18.27 g of zinc
nitrate dissolved into 100. ml of solution. 18.27 g 1 mol = mol Zn(NO3)2 189.4 g mol = M 0.100 L

7. How many grams of calcium phosphate are
needed to make 250. ml of M solution? L x mol = mol Ca3(PO4)2 L mol x g = 2.0 g Ca3(PO4)2 mol

HNO3 + NaOH → H2O + NaNO3 ? M 0.80 M 25.0 ml 15.0 ml
8. A 25.0 ml sample of nitric acid is titrated with 0.80 M sodium hydroxide. If neutralization requires 15.0 ml of the base, what is the molarity of the acid? HNO3 + NaOH → H2O NaNO3   ? M M 25.0 ml ml L x 0.80 mole/L = moles NaOH 1 mol NaOH = mol NaOH 1 mole HNO X X = mol HNO3  0.012 mol = M HNO3 L

3 H2SO4 + 2 Al(OH)3 → 6 H2O + Al2(SO4)3 0.80 M ? M 8.3 ml 10.0 ml
9. A 10.0 ml sample of aluminum hydroxide is titrated with 0.80 M sulfuric acid. If neutralization takes 8.3 ml of the acid, what is the molarity of the base? 3 H2SO Al(OH)3 → 6 H2O Al2(SO4)3   M ? M 8.3 ml ml L x 0.80 mole/L = moles H2SO4 2 mol Al(OH)3 = X = moles Al(OH)3 3 mole H2SO moles H2SO4 moles Al(OH)3 = M Al(OH)3 L

2 H3PO4 + 3 Ba(OH)2 → 6 H2O + Ba3(PO4)2 1.6 M ? M 4.2 ml 10.0 ml
10. A 10.0 ml sample of barium hydroxide is titrated with 1.6 M phosphoric acid. If neutralization takes 4.2 ml of the acid, what is the molarity of the base? 2 H3PO Ba(OH)2 → 6 H2O Ba3(PO4)2   M ? M 4.2 ml ml L x 1.6 mole/L = moles H3PO4 3 mol Ba(OH)2 = X = moles Ba(OH)2 2 mole H3PO moles H3PO4 0.010 moles Ba(OH)2 = 1.0 M Ba(OH)2 L