Question

1. calculate the ph of a solution with $\text{h}^+ = 6.0 \times 10^{-5} \text{ m}$.\
2. calculate the ph of a solution where $\text{oh}^- = 2.0 \times 10^{-3} \text{ m}$.\
3. a solution has poh = 3.20. what is its ph?\
4. what is the new ph when 500 ml of 0.10 m hcl is diluted to 2.0 l?\
5. a ph 3.00 solution is diluted until its ph becomes 4.00. by what factor does the hydrogen ion concentration change?\
6. a solution has a ph of 11.0. after dilution, its poh becomes... (partially visible) more acidic or more basic? explain quantitatively.

Explanation:

Problem 4: Calculate pH from [H⁺]

Step1: Use pH formula

pH = -log₁₀[H⁺] = -log₁₀(6.0×10⁻⁵)

Step2: Compute value

-log₁₀(6.0) ≈ 0.778, so pH ≈ 5 - 0.778 = 4.222

Problem 5: Calculate pH from [OH⁻]

Step1: Find pOH first

pOH = -log₁₀[OH⁻] = -log₁₀(2.0×10⁻³) ≈ 2.699

Step2: Relate pH and pOH

pH = 14 - pOH ≈ 14 - 2.699 = 11.301

Problem 6: pH from pOH

Step1: Apply pH + pOH =14

pH = 14 - 3.20 = 10.80

Problem7: pH after dilution of HCl

Step1: Find new [H⁺]

M₁V₁ = M₂V₂ → M₂ = (0.10×0.5)/2.0 = 0.025 M

Step2: Calculate pH

pH = -log₁₀(0.025) ≈ 1.602

Problem8: H⁺ concentration change

Step1: Find initial/final [H⁺]

Initial [H⁺] = 10⁻³, final [H⁺] =10⁻⁴

Step2: Compute ratio

Change factor = 10⁻⁴ /10⁻³ = 0.1 (1/10)

Problem9: Acidity change after dilution

Step1: Initial pOH

Initial pH=11 → initial pOH=14-11=3

Step2: Compare pOH (final missing, assume final pOH >3)

If final pOH >3 → final [OH⁻] =10⁻pOH <10⁻³ → less basic/more acidic

Answer:

1. ~4.22
2. ~11.30
3. 10.80
4. ~1.60
5. Decreases by a factor of 10
6. More acidic (if final pOH >3, [OH⁻] decreases)