Solved Titrations Practice Worksheet-Fxtra practice Find the ... - Free Printable

Let’s solve each problem step by step. We’ll use the balanced equations and titration formula:

Titration Formula (for 1:1 mole ratio):
M₁V₁ = M₂V₂
Where:
M = molarity (mol/L)
V = volume in mL or L (as long as units match on both sides)

But if the mole ratio is not 1:1, we must adjust using stoichiometry.

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Problem 1:

Write a balanced equation for NaOH + HCl → ?

NaOH + HCl → H₂O + NaCl
✔ Already balanced. Coefficients are all 1.

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Problem 2:

54 mL of 0.1 M NaOH neutralizes 125 mL HCl. Find [HCl].

From Problem 1: 1 mol NaOH reacts with 1 mol HCl → 1:1 ratio.

Use: M₁V₁ = M₂V₂
(0.1 M)(54 mL) = (M_HCl)(125 mL)
→ 5.4 = 125 × M_HCl
→ M_HCl = 5.4 / 125 = 0.0432 M

Check: 0.1 * 54 = 5.4 mmol NaOH → same mmol HCl → 5.4 mmol / 125 mL = 0.0432 M ✔

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Problem 3:

25 mL of 0.05 M HCl neutralizes 345 mL NaOH. Find [NaOH].

Again, 1:1 ratio from Problem 1.

M₁V₁ = M₂V₂
(0.05 M)(25 mL) = (M_NaOH)(345 mL)
→ 1.25 = 345 × M_NaOH
→ M_NaOH = 1.25 / 345 ≈ 0.00362 M

Check: 0.05 * 25 = 1.25 mmol HCl → same mmol NaOH → 1.25 / 345 ≈ 0.00362 M ✔

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Problem 4:

25.0 mL HCl titrated with 15.0 mL of 2.0 M NaOH. How many moles of HCl?

First, find moles of NaOH used:
moles NaOH = M × V(in L) = 2.0 mol/L × 0.015 L = 0.030 moles

Since 1:1 ratio, moles HCl = moles NaOH = 0.030 moles

Note: The question asks for moles in the 25 mL sample — that’s exactly what reacted, so answer is 0.030 moles.

✔ No need to find concentration unless asked.

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Problem 5:

Balanced equation: Ca(OH)₂ + HNO₃ → ?

Calcium hydroxide has 2 OH⁻, nitric acid has 1 H → need 2 HNO₃ per Ca(OH)₂.

Ca(OH)₂ + 2HNO₃ → 2H₂O + Ca(NO₃)₂

✔ Balanced.

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Problem 6:

50.0 mL of 0.5 M Ca(OH)₂ neutralizes 125 mL HNO₃. Find [HNO₃].

From Problem 5: 1 mol Ca(OH)₂ reacts with 2 mol HNO₃.

So, moles Ca(OH)₂ = 0.5 M × 0.050 L = 0.025 moles
→ Moles HNO₃ needed = 2 × 0.025 = 0.050 moles

Volume HNO₃ = 125 mL = 0.125 L
→ [HNO₃] = moles / volume = 0.050 / 0.125 = 0.4 M

Alternative way using modified titration formula:

For non-1:1 ratios:
(M_a × V_a × n_b) = (M_b × V_b × n_a)
Where n = coefficient from balanced equation.

Here:
Acid = HNO₃ (n=2), Base = Ca(OH)₂ (n=1)

So:
M_HNO₃ × V_HNO₃ × 1 = M_Ca(OH)₂ × V_Ca(OH)₂ × 2
Wait — better to think:

Moles base × its “OH count” = moles acid × its “H count”

Actually, standard method:

Moles Ca(OH)₂ = 0.5 × 0.050 = 0.025 mol
Each provides 2 OH → total OH⁻ = 0.050 mol
Each HNO₃ provides 1 H⁺ → need 0.050 mol HNO₃
In 0.125 L → 0.050 / 0.125 = 0.4 M ✔

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Problem 7:

How many mL of 0.50 M HNO₃ to titrate 25.0 mL of 0.05 M Ca(OH)₂?

From Problem 5: 1 Ca(OH)₂ : 2 HNO₃

Moles Ca(OH)₂ = 0.05 M × 0.025 L = 0.00125 mol
→ Moles HNO needed = 2 × 0.00125 = 0.0025 mol

Concentration HNO₃ = 0.50 M → Volume = moles / M = 0.0025 / 0.50 = 0.005 L = 5.0 mL

Check:
Using formula:
M_acid × V_acid × (coeff_base) = M_base × V_base × (coeff_acid)
Better:
Moles H⁺ needed = moles OH⁻ available
OH⁻ from Ca(OH)₂: 0.05 mol/L × 0.025 L × 2 = 0.0025 mol OH⁻
→ Need 0.0025 mol H⁺ → from 0.50 M HNO → V = 0.0025 / 0.50 = 0.005 L = 5.0 mL ✔

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Problem 8:

75.0 mL of 1.5 M HNO₃ neutralizes 125 mL Ca(OH)₂. Find moles of Ca(OH)₂.

From Problem 5: 2 HNO₃ : 1 Ca(OH)₂

Moles HNO₃ = 1.5 M × 0.075 L = 0.1125 mol
→ Moles Ca(OH)₂ = 0.1125 / 2 = 0.05625 moles

Check:
Each Ca(OH)₂ takes 2 HNO₃ → so half the moles of acid → 0.1125 / 2 = 0.05625 mol ✔

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Final Answers:

1) NaOH + HCl → H₂O + NaCl
2) 0.0432 M
3) 0.00362 M
4) 0.030 moles
5) Ca(OH)₂ + 2HNO₃ → 2H₂O + Ca(NO₃)₂
6) 0.4 M
7) 5.0 mL
8) 0.05625 moles