Printable worksheet

← Back to the pH Calculator

10 problems20 problems

Include answer key

In the print window, choose “Save as PDF” as the printer (or print it on paper). Seed 1 rebuilds this exact sheet.

pH worksheet

Grades 10–12 · CCSS HSF.BF.B.5

Solve each problem and show your work. Round answers to 2 decimal places where needed.

1.
Problem 1.
- What do you know?: pOH
- pOH: 4
Find: pH
Answer
2.
Problem 2.
- What do you know?: Hydroxide ion concentration [OH⁻]
- [OH⁻]: 0.002 mol/L
Find: pH
Answer
3.
Problem 3.
- What do you know?: pH
- pH: 3
Find: Hydrogen ion concentration [H⁺]
Answer
4.
Problem 4.
- What do you know?: Hydrogen ion concentration [H⁺]
- [H⁺]: 0.0002 mol/L
Find: pH
Answer
5.
Problem 5.
- What do you know?: Hydroxide ion concentration [OH⁻]
- [OH⁻]: 0.001 mol/L
Find: pH
Answer
6.
Problem 6.
- What do you know?: Strong acid concentration
- Concentration: 0.01 mol/L
- H⁺ or OH⁻ released per formula unit: 1
Find: pH
Answer
7.
Problem 7.
- What do you know?: Strong acid concentration
- Concentration: 0.01 mol/L
- H⁺ or OH⁻ released per formula unit: 2
Find: pH
Answer
8.
Problem 8.
- What do you know?: Strong base concentration
- Concentration: 0.01 mol/L
- H⁺ or OH⁻ released per formula unit: 2
Find: pH
Answer
9.
Problem 9.
- What do you know?: Hydrogen ion concentration [H⁺]
- [H⁺]: 0.00028 mol/L
Find: pH
Answer
10.
Problem 10.
- What do you know?: pOH
- pOH: 3.6
Find: pH
Answer

Answer key: pH worksheet

1. pH: 10
1. Find pH: at 25 °C, pH + pOH always equals 14, so subtract pOH from 14
  $pH$ $= 14 - 4$ $= 10$
2. Undo the logarithm to get the hydrogen ion concentration: raise 10 to the power −pH
  $[H⁺]$ $= 10^{−10}$ $= 1 × 10⁻¹⁰ mol/L$
3. Undo the logarithm to get the hydroxide concentration: raise 10 to the power −pOH
  $[OH⁻]$ $= 10^{−4}$ $= 1 × 10⁻⁴ mol/L$
2. pH: 11.3
1. Find pOH: take the base-10 logarithm of the hydroxide ion concentration (its power of ten) and flip the sign
  $pOH$ $= −log₁₀ (0.002)$ $= 2.69897$
2. Find pH: at 25 °C, pH + pOH always equals 14, so subtract pOH from 14
  $pH$ $= 14 - 2.69897$ $= 11.30103$
3. Find the hydrogen ion concentration. In water, [H⁺] × [OH⁻] always equals Kw, so divide Kw by [OH⁻]
  $[H⁺]$ $= \frac{1 × 10⁻¹⁴}{0.002}$ $= 5 × 10⁻¹² mol/L$
3. Hydrogen ion concentration [H⁺]: 0.001 mol/L
1. Undo the logarithm to get the hydrogen ion concentration: raise 10 to the power −pH
  $[H⁺]$ $= 10^{−3}$ $= 0.001 mol/L$
2. Find pOH: at 25 °C, pH + pOH always equals 14, so subtract pH from 14
  $pOH$ $= 14 - 3$ $= 11$
3. Undo the logarithm to get the hydroxide concentration: raise 10 to the power −pOH
  $[OH⁻]$ $= 10^{−11}$ $= 1 × 10⁻¹¹ mol/L$
4. pH: 3.7
1. Find pH: take the base-10 logarithm of the hydrogen ion concentration (its power of ten) and flip the sign
  $pH$ $= −log₁₀ (2 × 10⁻⁴)$ $= 3.69897$
2. Find pOH: at 25 °C, pH + pOH always equals 14, so subtract pH from 14
  $pOH$ $= 14 - 3.69897$ $= 10.30103$
3. Find the hydroxide concentration. In water, [H⁺] × [OH⁻] always equals Kw, so divide Kw by [H⁺]
  $[OH⁻]$ $= \frac{1 × 10⁻¹⁴}{2 × 10⁻⁴}$ $= 5 × 10⁻¹¹ mol/L$
5. pH: 11
1. Find pOH: take the base-10 logarithm of the hydroxide ion concentration (its power of ten) and flip the sign
  $pOH$ $= −log₁₀ (0.001)$ $= 3$
2. Find pH: at 25 °C, pH + pOH always equals 14, so subtract pOH from 14
  $pH$ $= 14 - 3$ $= 11$
3. Find the hydrogen ion concentration. In water, [H⁺] × [OH⁻] always equals Kw, so divide Kw by [OH⁻]
  $[H⁺]$ $= \frac{1 × 10⁻¹⁴}{0.001}$ $= 1 × 10⁻¹¹ mol/L$
6. pH: 2
1. A strong acid breaks apart completely, so [H⁺] equals its ion concentration. The water's own ions are too few to matter here
  $[H⁺]$ $= 0.01 mol/L$
2. Find pH: take the base-10 logarithm of the hydrogen ion concentration (its power of ten) and flip the sign
  $pH$ $= −log₁₀ (0.01)$ $= 2$
3. Find pOH: at 25 °C, pH + pOH always equals 14, so subtract pH from 14
  $pOH$ $= 14 - 2$ $= 12$
4. Find the hydroxide concentration. In water, [H⁺] × [OH⁻] always equals Kw, so divide Kw by [H⁺]
  $[OH⁻]$ $= \frac{1 × 10⁻¹⁴}{0.01}$ $= 1 × 10⁻¹² mol/L$
7. pH: 1.7
1. Each formula unit releases 2 ions, so multiply the concentration by 2 to get the ion concentration
  $2 \times 0.01 mol/L$ $= 0.02 mol/L$
2. A strong acid breaks apart completely, so [H⁺] equals its ion concentration. The water's own ions are too few to matter here
  $[H⁺]$ $= 0.02 mol/L$
3. Find pH: take the base-10 logarithm of the hydrogen ion concentration (its power of ten) and flip the sign
  $pH$ $= −log₁₀ (0.02)$ $= 1.69897$
4. Find pOH: at 25 °C, pH + pOH always equals 14, so subtract pH from 14
  $pOH$ $= 14 - 1.69897$ $= 12.30103$
5. Find the hydroxide concentration. In water, [H⁺] × [OH⁻] always equals Kw, so divide Kw by [H⁺]
  $[OH⁻]$ $= \frac{1 × 10⁻¹⁴}{0.02}$ $= 5 × 10⁻¹³ mol/L$
8. pH: 12.3
1. Each formula unit releases 2 ions, so multiply the concentration by 2 to get the ion concentration
  $2 \times 0.01 mol/L$ $= 0.02 mol/L$
2. A strong base breaks apart completely, so [OH⁻] equals its ion concentration. The water's own ions are too few to matter here
  $[OH⁻]$ $= 0.02 mol/L$
3. Find pOH: take the base-10 logarithm of the hydroxide ion concentration (its power of ten) and flip the sign
  $pOH$ $= −log₁₀ (0.02)$ $= 1.69897$
4. Find pH: at 25 °C, pH + pOH always equals 14, so subtract pOH from 14
  $pH$ $= 14 - 1.69897$ $= 12.30103$
5. Find the hydrogen ion concentration. In water, [H⁺] × [OH⁻] always equals Kw, so divide Kw by [OH⁻]
  $[H⁺]$ $= \frac{1 × 10⁻¹⁴}{0.02}$ $= 5 × 10⁻¹³ mol/L$
9. pH: 3.55
1. Find pH: take the base-10 logarithm of the hydrogen ion concentration (its power of ten) and flip the sign
  $pH$ $= −log₁₀ (2.8 × 10⁻⁴)$ $= 3.552842$
2. Find pOH: at 25 °C, pH + pOH always equals 14, so subtract pH from 14
  $pOH$ $= 14 - 3.552842$ $= 10.447158$
3. Find the hydroxide concentration. In water, [H⁺] × [OH⁻] always equals Kw, so divide Kw by [H⁺]
  $[OH⁻]$ $= \frac{1 × 10⁻¹⁴}{2.8 × 10⁻⁴}$ $= 3.571 × 10⁻¹¹ mol/L$
10. pH: 10.4
1. Find pH: at 25 °C, pH + pOH always equals 14, so subtract pOH from 14
  $pH$ $= 14 - 3.6$ $= 10.4$
2. Undo the logarithm to get the hydrogen ion concentration: raise 10 to the power −pH
  $[H⁺]$ $= 10^{−10.4}$ $= 3.981 × 10⁻¹¹ mol/L$
3. Undo the logarithm to get the hydroxide concentration: raise 10 to the power −pOH
  $[OH⁻]$ $= 10^{−3.6}$ $= 2.512 × 10⁻⁴ mol/L$