Question

If 23.7 mL of .14 M sodium hydroxide neutralizes 5.0 mL of vinegar, what is the concentration of the vinegar?

Answer 1

1. If 10.0 mL of 0.100 M HCl is titrated with 0.200 M NaOH, what volume of sodium
hydroxide solution is required to neutralize the acid?
HCl(aq) + NaOH(aq)  NaCl(aq) + H2O(l)
M1V1 = M2V2 (0.100M) (0.010L) = (0.200M)(V2) V2 = 0.005 L = 5 mL
2. If 20.0 mL of 0.500 M KOH is titrated with 0.250 M HNO3, what volume of nitric
acid is required to neutralize the base?
HNO3(aq) + KOH(aq)  KNO3(aq) + H2O(l)
M1V1 = M2V2 (0.500M) (0.020L) = (0.250M)(V2) V2 = 0.040 L = 40 mL
3. If 25.0 mL of 0.100 M HCl is titrated with 0.150 M Ba(OH)2, what volume of
barium hydroxide is required to neutralize the acid?
2 HCl(aq) + Ba(OH)2(aq)  BaCl2(aq) + 2 H2O(l)
M1V1 = M2V2 (0.100M) (0.025L) = (0.150M)(V2) V2 = 0.0166 L = 16.6 mL OH-
But there are 2 OH’s per Ba(OH)2 so it takes half this volume = 8.33 mL of Ba(OH)2
4. If 25.0 mL of 0.100 M Ca(OH)2 is titrated with 0.200 M HNO3, what volume of
nitric acid is required to neutralize the base?
2 HNO3(aq) + Ca(OH)2(aq)  2 Ca(NO3)2(aq) + 2 H2O(l)
M1V1 = M2V2 (0.100M) (0.025L) = (0.200M)(V2) V2 = 0.0125 L = 12.5 mL H+
But it takes 2 HNO3’s per Ca(OH)2 so it takes twice this volume = 25 mL of HNO3
5. If 20.0 mL of 0.200 M H2SO4 is titrated with 0.100 M NaOH, what volume of
sodium hydroxide is required to neutralize the acid?
H2SO4(aq) + 2 NaOH(aq)  Na2SO4(aq) + 2 H2O(l)
0.200 M H2SO4 = 0.400 M H+
M1V1 = M2V2 (0.40M) (0.020L) = (0.100M)(V2) V2 = 0.080 L = 80 mL NaOH
6. If 30.0 mL of 0.100 M Ca(OH)2 is titrated with 0.150 M HC2H3O2, what volume of
acetic acid is required to neutralize the base?
2 HC2H3O2(aq) + Ca(OH)2(aq)  Ca(C2H3O2)2(aq) + 2 H2O(l)
0.100 M Ca(OH)2 = 0.200 M OH-
M1V1 = M2V2 (0.200M) (0.030L) = (0.150M)(V2) V2 = 0.040 L = 40 mL NaOH
7. If a 50.0 mL sample of ammonium hydroxide is titrated with 25.0 mL of 0.200 M
nitric acid to a methyl red endpoint, what is the molarity of the base?
NH4OH(aq) + HNO3(aq)  NH4NO3(aq) + H2O(l)
M1V1 = M2V2 (0.200M) (0.025L) = (M2)(0.050L) M2 = 0.100 M NH4OH
8. If a 50.0 mL sample of ammonium hydroxide is titrated with 25.0 mL of 0.200 M
sulfuric acid to a methyl red endpoint, what is the molarity of the base?
2 NH4OH(aq) + H2SO4(aq)  (NH4)2SO4(aq) + 2 H2O(l)
0.200 M H2SO4 = 0.400 M H+
M1V1 = M2V2 (0.400M) (0.025L) = (M2)(0.050L) M2 = 0.200 M NH4OH
9. If a 25.0 mL sample of sulfuric acid is titrated with 50.0 mL of 0.200 M potassium
hydroxide to a phenolphthalein endpoint, what is the molarity of the acid?
H2SO4(aq) + 2 KOH(aq)  K2SO4(aq) + 2 H2O(l)
M1V1 = M2V2 (0.200M) (0.050L) = (M2)(0.025L) M2 = 0.400 M H+
But, there are 2 H’s per H2SO4 so [H2SO4] = 0.200M
10. What is the molarity of a hydrochloric acid solution if 20.00 mL of HCl is required
to neutralize 0.424 g of sodium carbonate (105.99 g/mol)?
2 HCl(aq) + Na2CO3(aq)  2 NaCl(aq) + H2O(l) + CO2(g)
0.424 g/105.99 g/mol = 0.0040 mol Na2CO3
Each Na2CO3 requires 2 HCl so we need 0.0080 mol HCl
MV = moles (M)(0.020L) = 0.0080 mole HCl M = 0.40 M HCl
11. What is the molarity of a nitric acid solution if 25.00 mL of HNO3 is required to
neutralize 0.424 g of sodium carbonate (105.99 g/mol)?
2 HNO3(aq) + Na2CO3(aq)  2 NaNO3(aq) + H2O(l) + CO2(g)
0.424 g/105.99 g/mol = 0.0040 mol Na2CO3
Each Na2CO3 requires 2 HNO3 so we need 0.0080 mol HNO3
MV = moles (M)(0.025L) = 0.0080 mole HNO3 M = 0.32 M HNO3
12. What is the molarity of a sulfuric acid solution if 30.00 mL of H2SO4 is required to
neutralize 0.840 g of sodium hydrogen carbonate (84.01 g/mol)?
H2SO4(aq) + 2 NaHCO3(aq)  Na2SO4(aq) + 2 H2O(l) + 2 CO2(g)
0.840 g / 84.01 g/mol = 0.010 mol NaHCO3
It takes 2 NaHCO3 per H2SO4 so you need 0.005 mol H2SO4
MV = moles M(0.030L) = 0.005 moles M = 0.167 M H2SO4
13. What is the molarity of a hydrochloric acid solution if 25.00 mL of HCl is required
to neutralize 0.500 g of calcium carbonate (100.09 g/mol)?
2 HCl(aq) + CaCO3(s)  CaCl2(aq) + H2O(l) + CO2(g)
0.500 g/100.09 g/mol = 0.005 mol CaCO3
Each mole of CaCO3 requires 2 mol HCl so you need 0.005 x 2 = 0.010 mol HCl
MV = moles M(0.025L) = 0.010 mol M = 0.40 M HCl
14. What is the molarity of a sodium hydroxide solution if 40.00 mL of NaOH is
required to neutralize 0.900 g of oxalic acid, H2C2O4, (90.04 g/mol)?
H2C2O4(aq) + 2 NaOH(aq) Na2C2O4(aq) + 2 H2O(l)
0.900 g / 90.04 g/mol = 0.010 mol Oxalic acid
It takes 2 mole NaOH for every mole of Oxalic acid
so you need 2 x 0.010 mol = 0.02 mol NaOH
MV = moles M(0.040L) = 0.020 mole NaOH M = 0.50 M NaOH