Problem #1: How many mL of 1.80 M hydrochloric acid are required to completly react with 30.0 g of calcium hydroxide?
Solution:
1) Moles of Ca(OH)2:
30.0 g / 74.0918 g/mol = 0.4049031 mol
2) Chemical equation for the reaction:
2HCl + Ca(OH)2 ---> CaCl2 + 2H2OThe molar ratio between HCl and calcium hydroxide is 2:1
Two moles of HCl are used up for every one mole of Ca(OH)2 reacted.
3) Moles of HCl used:
0.4049031 mol times 2 = 0.8098062 mol
4) Volume of HCl required:
moles = MV0.8098062 mol = (1.80 mol/L) (x)
x = 0.449892 L
x = 450. mL (to three sig figs, note the explicit decimal point)
Problem #2: How many milliliters of 0.122 M HCl would be required to titrate 6.45 g Ba(OH)2?
Solution:
1) The balanced equation for the reaction:
2HCl(aq) + Ba(OH)2(s) ---> BaCl2(aq) + 2H2O(ℓ)
2) Moles of Ba(OH)2:
6.45 g / 171.3438 g/mol = 0.0376436 mol
3) There is a 2 to 1 molar ratio between HCl and Ba(OH)2. Consequently:
(0.0376436 mol) (2) = 0.0752872 mole of HCl used
4) Volume of HCl required:
0.0752872 mole / 0.122 mol/L = 0.617 L = 617 mL
Note that MV = mass / molar mass was not used. MV = mass / molar mass only works when there is a 1:1 molar ratio.
Problem #3: What is the molarity of your HCl solution if it takes 35.25 mL to reach the second equivalence point of a titration against 0.2225 grams of standard sodium carbonate?
Solution #1:
Reaching the second equivalence point means this reaction:2HCl(aq) + Na2CO3(s) ---> 2NaCl(aq) + CO2(g) + H2O(ℓ)That's a 2:1 ratio: two moles of HCl are needed to react with one mole of sodium carbonate.
0.2225 g / 105.988 g/mol = 0.0020993 mol of Na2CO3
(0.0020993 mol) (2) = 0.0041986 mol of HCl required
0.0041986 mol / 0.03525 L = 0.1191 M
Solution #2:
Consider this reaction:HCl(aq) + Na2CO3(s) ---> NaHCO3(aq) + NaCl(aq)That's the reaction for the first equivalence point.
35.25 mL divided by 2 to get to first eq. point ---> 17.625 mL
0.2225 g / 105.988 g/mol = 0.0020993 mol of sodium carbonate
HCl and Na2CO3 react in a 1 to 1 molar ratio to produce the NaHCO3
0.0020993 mol of HCl was required.
0.0020993 mol / 0.017625 L = 0.1191 M
Problem #4: How many milliliters of 0.200 M HCl can react with 6.25 g CaCO3?
Solution:
CaCO3(s) + 2HCl(aq) ---> CaCl2(aq) + CO2(g) + H2O(ℓ)Moles CaCO3 ---> 6.25 g / 100.086 g/mol = 0.0624463 mol
From the balanced equation, 2 moles of HCl are needed for every mole of CaCO3.
Moles HCl ---> 0.0624463 mol times 2 = 0.1248926 mol
Volume HCl required ---> 0.1248926 divided by 0.200 mol/L = 0.624463 L
624 mL (to three sig figs)
Problem #5: What is the molarity of sulfuric acid if 43.90 mL of H2SO4 is required to neutralize 0.455 g of sodium hydrogen carbonate?
Solution:
1) The reaction is:
H2SO4(aq) + 2NaHCO3(aq) ---> Na2SO4(aq) + 2CO2(g) + 2H2O(ℓ)The key ratio is that 1 mole of sulfuric acid is needed for every 2 moles of NaHCO3.
2) Moles of NaHCO3:
0.455 g / 84.0059 g/mol = 0.0054163 mol
3) Moles H2SO4 required:
0.0054163 mol / 2 = 0.00270815 molThe fact that I divided by two flows out of how the problem was worded. Usually, it's worded so as to produce a multiplication by two. Not in this problem.
4) Molarity:
0.00270815 mol / 0.04390 L = 0.0617 M
Problem #6: What is the molarity of nitric acid if 42.60 mL of HNO3 is required to neutralize 0.555 g of sodium carbonate?
Solution:
1) The reaction is:
2HNO3(aq) + Na2CO3(aq) ---> 2NaNO3(aq) + CO2 + H2O(ℓ)The key molar ratio is that 2 moles of nitric acid are required for every one mole of sodium carbonate neutralized.
2) Moles of Na2CO3:
0.555 g / 105.988 g/mol = 0.00523644 mol
3) Moles nitric acid required:
(0.00523644 mol) (2) = 0.01047288 mol
4) Molarity:
0.01047288 mol / 0.04260 L = 0.246 M
Problem #7: What is the molarity of sodium hydroxide if 35.60 mL of NaOH is required to neutralize 0.631 g of oxalic acid, H2C2O4?
Solution #1:
1) The reaction is:
H2C2O4(aq) + 2NaOH(aq) ---> Na2C2O4(aq) + 2H2O(ℓ)The key is the 1 to 2 molar ratio between oxalic acid and sodium hydroxide.
2) Moles of oxalic acid:
0.631 g / 90.0338 g/mol = 0.0070085 mol
3) Moles of NaOH required:
(0.0070085 mol) (2) = 0.014017 mol
4) Molarity:
0.014017 mol / 0.03560 L = 0.394 M
Solution #2:
This solution is based on the idea that it takes half the volume of sodium hydroxide solution to reach the first equivalence point.
1) The reaction to the first equivalence point is:
H2C2O4(aq) + NaOH(aq) ---> NaHC2O4(aq) + H2O(ℓ)The key idea is that the molar ratio between the reactants is a one to one ratio.
2) Moles of oxalic acid:
0.631 g / 90.0338 g/mol = 0.0070085 mol
3) Moles of NaOH required:
0.0070085 mol (based on 1:1 ratio)
4) Molarity:
0.0070085 mol / 0.0178 L = 0.394 MNote the use of a volume exactly one-half of 35.60 mL. That is because we only titrated to the first equivalence point.
Problem #8: What volume of 0.189 M barium hydroxide is required to neutralize 1.228 g of potassium hydrogen phthalate, KHC8H4O4 (204.23 g/mol)?
Solution:
1) The reaction is:
2KHC8H4O4(aq) + Ba(OH)2(aq) ---> BaC8H4O4(s) + K2C8H4O4(aq) + 2H2O(ℓ)The key ratio is the 2 to 1 molar ratio between KHP and barium hydroxide.
2) moles KHP:
1.228 g / 204.23 g/mol = 0.006013 mol
3) moles barium hydroxide:
0.006013 mol / 2 = 0.0030065 mol
4) volume required:
0.189 mol/L = 0.0030065 mol / xx = 0.0159 L (or 15.9 mL)
Problem #9: What volume of 0.765 M H3PO4 is required to exactly neutralize 2.000 g of calcium hydroxide?
Solution:
2.000 g / 74.0918 g/mol = 0.02699354 mol of Ca(OH)22H3PO4(aq) + 3Ca(OH)2(s) ---> Ca3(PO4)2(s) + 6H2O(ℓ)
Two moles of H3PO4 are required for every three moles of Ca(OH)2
2 is to 3 as x is to 0.02699354 mol
x = 0.01799569 mol of H3PO4 required
0.01799569 mol divided by 0.765 mol/L = 0.0235 L = 23.5 mL
Problem #10: A 2.61 M solution of phosphoric acid (H3PO4) is to be reacted with aluminum hydroxide to make aluminum phosphate and water. How many mL of the phosphoric acid solution are needed to react with 26.7 g of aluminum hydroxide?
Solution:
H3PO4(aq) + Al(OH)3(s) ---> AlPO4(s) + 3H2O(ℓ)One mole of H3PO4 is required for every one mole of Al(OH)3
26.7 g / 78.00367 g/mol = 0.342292 mol of Al(OH)3
Due to 1 to 1 molar ratio between the reactants, 0.342292 mol of H3PO4 reacted.
0.342292 mol / 2.61 mol/L = 0.131 L = 131 mL (to three sig figs)
Bonus Problem #1: An artificial fruit beverage contains 12.00 g of tartaric acid (a diprotic acid), H2C4H4O6, to achieve the tartness. It is titrated with a basic solution that has a density of 1.045 g/cm3 and contains 5.000 mass percent KOH. What volume of the basic solution is required?
Solution:
1) We need to know the moles of the acid:
12.00 g / 150.0854 g/mol = 0.0799545 mol
2) The acid is diprotic, so we know we need twice as many moles of the base as moles of acid. Here's the chemical equation (H2T = tartaric acid):
H2T + 2KOH ---> K2T + 2H2O0.0799545 mol x 2 = 0.159909 mol of KOH needed
3) Since there is a mass percent for solution concentration, let's convert to grams of KOH:
0.159909 mol times 56.1049 g/mol = 8.97168 g
4) What mass of 5% solution will deliver 8.97168 g of KOH?
8.97168 g / 0.05 = 179.4336 g
5) What is the volume of 179.4336 g of 5% KOH solution?
179.4336 g / 1.045 g/cm3 = 171.7068 cm3Four significant figures yields 171.7 cm3 for the answer.
Bonus Problem #2: A student wanted to determine the percent by mass of NaCl in a mixture (3.000 g) which contained NaCl & KHP. He was given some KOH solution with unknown concentration & some 0.2470 M H2SO4 . What should he do?
Discussion:
For those of you that do not know, KHP stands for potassium hydrogen phthalate, C8H5KO4. A fuller name for it is 1,2-benzenedicarboxylic acid, monopotassium salt. It is in wide use as a primary standard in the acid base world and using KHP is in wide use.Just to slam dunk it, make sure you realize that the P stands for phthalate, not phosphorous.
Also, the 20.00 mL and 1.500 g were simply chosen by the question writer. zHe/She could have used any set of numbers he/she desired
Solution:
1) He titrated 20.00 mL of H2SO4 and found that it took 33.70 mL of KOH. Calculate the molarity of the KOH solution.
H2SO4 + 2KOH ---> K2SO4 + 2H2Omillimoles H2SO4 ---> (0.2470 mmol/mL) ( 20.00 mL) = 4.94 mmol
H2SO4 and KOH react in a 1:2 molar ratio
mmole of KOH consumed ---> (4.940 mmol) (2) = 9.880 mmol
molarity of KOH solution ---> 9.880 mmol / 33.70 mL = 0.293175 M (keep some guard digits)
2) Titrate 1.500 g of the mixture with KOH, finding that 11.50 mL of KOH were required to reach the end point. Determine mass of KHP that reacts.
KHP + KOH ---> K2P +H2Ommoles KOH ---> (0.293175 mmol/mL) (11.50 mL) = 3.3715125 mmol
KOH and KHP react in a 1:1 molar ratio, therefore 3.3715125 mmol of KHP was consumed.
3.3715125 mmol = 0.0033715125 mol
(204.2215 g/mol) (0.0033715125 mol) = 0.68853534 g
3) Determine mass percent of NaCl:
0.68853534 g / 1.500 g = 0.4590 (decimal percent of KHP)1 − 0.4590 = 0.5410 (decimal percent of NaCl)
NaCl is 54.10%
The other way would be to subtract first to get the mass of NaCl. Then, you divide by 1.500 to get the NaCl decimal percent. Multiply by 100 gives the percentage.