IMPORTANT NUMERICALS FROM VOLUMETRIC ANALYSIS |EXAM ORIENTED|CLASS 12|NEB
Volumetric analysis is a common analytical chemistry technique that involves measuring the volume of a solution of known concentration that is required to react completely with a solution of unknown concentration. This technique is widely used in various applications, including titration and acid-base analysis. Here are some examples of numerical questions related to volumetric analysis:
1.A 25.0 mL solution of hydrochloric acid of unknown concentration requires 38.2 mL of 0.100 M sodium hydroxide for complete neutralization. Find the concentration of the hydrochloric acid solution?
Solution:
HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)
From the equation, we can see that one mole of HCl reacts with one mole of NaOH. Therefore, the number of moles of NaOH used can be calculated as:
moles NaOH = concentration × volume = 0.100 M × 38.2 mL ÷ 1000 mL/L = 0.00382 moles
Since one mole of NaOH reacts with one mole of HCl, the number of moles of HCl present in the solution can be calculated as:
moles HCl = 0.00382 moles
The volume of the HCl solution used is 25.0 mL or 0.025 L. Therefore, the concentration of the HCl solution can be calculated as:
concentration = moles ÷ volume = 0.00382 moles ÷ 0.025 L = 0.153 M
Therefore, the concentration of the hydrochloric acid solution is 0.153 M.
2.A 50.0 mL sample of a sulfuric acid solution requires 45.2 mL of 0.100 M sodium hydroxide for complete neutralization. What is the concentration of the sulfuric acid solution?
Solution:
H2SO4(aq) + 2NaOH(aq) → Na2SO4(aq) + 2H2O(l)
From the equation, we can see that two moles of NaOH react with one mole of H2SO4. Therefore, the number of moles of NaOH used can be calculated as:
moles NaOH = concentration × volume = 0.100 M × 45.2 mL ÷ 1000 mL/L = 0.00452 moles
Since two moles of NaOH react with one mole of H2SO4, the number of moles of H2SO4 present in the solution can be calculated as:
moles H2SO4 = 0.00452 moles ÷ 2 = 0.00226 moles
The volume of the H2SO4 solution used is 50.0 mL or 0.050 L. Therefore, the concentration of the H2SO4 solution can be calculated as:
concentration = moles ÷ volume = 0.00226 moles ÷ 0.050 L = 0.0452 M
Therefore, the concentration of the sulfuric acid solution is 0.0452 M.
3.A 20.0 mL sample of an acetic acid solution requires 18.5 mL of 0.200 M sodium hydroxide for complete neutralization. What is the concentration of the acetic acid solution?
Solution:
CH3COOH(aq) + NaOH(aq) → NaCH3COO(aq) + H2O(l)
we can see that one mole of CH3COOH reacts withone mole of NaOH. Therefore, the number of moles of NaOH used can be calculated as:
moles NaOH = concentration × volume = 0.200 M × 18.5 mL ÷ 1000 mL/L = 0.00370 moles
Since one mole of NaOH reacts with one mole of CH3COOH, the number of moles of CH3COOH present in the solution can be calculated as:
moles CH3COOH = 0.00370 moles
The volume of the CH3COOH solution used is 20.0 mL or 0.020 L. the concentration of the CH3COOH solution can be calculated as:
concentration = moles ÷ volume = 0.00370 moles ÷ 0.020 L = 0.185 M
so, the concentration of the acetic acid solution is 0.185 M.
4.A 30.0 mL sample of a potassium hydroxide solution requires 25.6 mL of 0.150 M hydrochloric acid for complete neutralization. What is the concentration of the potassium hydroxide solution?
Solution:
KOH(aq) + HCl(aq) → KCl(aq) + H2O(l)
we can see that one mole of KOH reacts with one mole of HCl. the number of moles of HCl used can be calculated as:
moles HCl = concentration × volume = 0.150 M × 25.6 mL ÷ 1000 mL/L = 0.00384 moles
one mole of KOH reacts with one mole of HCl, so the number of moles of KOH present in the solution can be calculated as:
moles KOH = 0.00384 moles
The volume of the KOH solution used is 30.0 mL or 0.030 L. Therefore, the concentration of the KOH solution can be calculated as:
concentration = moles ÷ volume = 0.00384 moles ÷ 0.030 L = 0.128 M
Therefore, the concentration of the potassium hydroxide solution is 0.128 M.
5.A 40.0 mL sample of a nitric acid solution requires 35.0 mL of 0.100 M sodium hydroxide for complete neutralization. What is the concentration of the nitric acid solution?
Solution:
HNO3(aq) + NaOH(aq) → NaNO3(aq) + H2O(l)
moles NaOH = concentration × volume = 0.100 M × 35.0 mL ÷ 1000 mL/L = 0.00350 moles
moles HNO3 = 0.00350 moles
The volume of the HNO3 solution used is 40.0 mL or 0.040 L. Therefore, the concentration of the HNO3 solution can be calculated as:
concentration = moles ÷ volume = 0.00350 moles ÷ 0.040 L = 0.088 M
Therefore, the concentration of the nitric acid solution is 0.088 M.
6.A 25.0 mL sample of a hydrochloric acid solution requires 12.5 mL of 0.200 M sodium hydroxide for complete neutralization. What is the concentration of the hydrochloric acid solution?
Solution:
The balanced chemical equation for the reaction is:
HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)
moles NaOH = concentration × volume = 0.200 M × 12.5 mL ÷ 1000 mL/L = 0.00250 moles
moles HCl = 0.00250 moles
The volume of the HCl solution used is 25.0 mL or 0.025 L. Therefore, the concentration of the HCl solution can be calculated as:
concentration = moles ÷ volume = 0.00250 moles ÷ 0.025 L = 0.100 M
Therefore, the concentration of the hydrochloric acid solution is 0.100 M.
7.A 50.0 mL sample of a sulfuric acid solution requires 20.0 mL of 0.100 M sodium hydroxide for complete neutralization. What is the concentration of the sulfuric acid solution?
Solution:
H2SO4(aq) + 2NaOH(aq) → Na2SO4(aq) + 2H2O(l)
moles NaOH = concentration × volume = 0.100 M × 20.0 mL ÷ 1000 mL/L = 0.00200 moles
moles H2SO4 = 0.00200 moles ÷ 2 = 0.00100 moles
The volume of the H2SO4 solution used is 50.0 mL or 0.050 L. Therefore, the concentration of the H2SO4 solution can be calculated as:
concentration = moles ÷ volume = 0.00100 moles ÷ 0.050 L = 0.020 M
the concentration of the sulfuric acid solution is 0.020 M.
8.A 10.0 mL sample of a sodium hydroxide solution requires 12.5 mL of 0.100 M hydrochloric acid for complete neutralization. What is the concentration of the sodium hydroxide solution?
Solution:
NaOH(aq) + HCl(aq) → NaCl(aq) + H2O(l)
moles HCl = concentration × volume = 0.100 M × 12.5 mL ÷ 1000 mL/L = 0.00125 moles
moles NaOH = 0.00125 moles
The volume of the NaOH solution used is 10.0 mL or 0.010 L. Therefore, the concentration of the NaOH solution can be calculated as:
concentration = moles ÷ volume = 0.00125 moles ÷ 0.010 L = 0.125 M
Therefore, the concentration of the sodium hydroxide solution is 0.125 M.
These questions have been carefully selected to help you prepare for your upcoming exams and tests. However, I urge you not to rely solely on these questions, but to also practice solving problems and understanding the underlying concepts.
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