The pH of a 0.01 M weak acid HX(Kₐ=4 × 10⁻¹⁰) is found to be 5 . Now the acid solution is diluted with excess of water so that the pH of the solution changes to 6 . The new concentration of the diluted weak acid is given as x × 10⁻⁴ M. The value of x is _________ (nearest integer)

Solution

Chemistry ma’am · 3 steps

For a weak acid, [H⁺]² = Kₐ C. Put in the new pH and solve for the new concentration.

1. Weak acid formula. Kₐ = 4 × 10⁻¹⁰ is very small, so HX ionises only slightly and [H⁺] = √(Kₐ C). Kₐ does not change on dilution, so only the new pH is needed. The original 0.01 M and pH 5 do not enter the calculation.

   C = [H⁺]²/Kₐ

2. Put in numbers. pH 6 means [H⁺] = 10⁻⁶ M, so [H⁺]² = 10⁻¹². Then C = 10⁻¹²/(4 × 10⁻¹⁰) = 2.5 × 10⁻³ M.
3. So. C = 2.5 × 10⁻³ M = 25 × 10⁻⁴ M, so x = 25.

The common slip: 2.5. Stopping at 2.5 × 10⁻³ and reporting 2.5 gives the wrong x. The answer must be written as a multiple of 10⁻⁴, which is 25 × 10⁻⁴.

The idea to keep. Dilution changes the concentration but never Kₐ. For any weak acid, read [H⁺] from the pH, square it, and divide by Kₐ to get the concentration.

Answer from the NTA final answer key. Solution by PrepWiser.