Question

buffers are solutions that buffer against large changes in ph when an acid or base is added. buffers contain a weak acid to neutralize added base, and a weak base to neutralize added acid. most buffer problems want you to calculate the ph. for example, calculate the ph of a buffer made from dihydrogen phosphate and hydrogen phosphate when the h₂po₄⁻¹ = 4.44 × 10⁻⁵ m, the hpo₄⁻² = 5.55 × 10⁻⁵ m, and the kₐ = 6.20 × 10⁻⁸. i. first, identify the acid. its either h₂po₄⁻¹ or hpo₄⁻². the acid is 1 because it has the most acidic hydrogens. ii. then write the reaction for the acid dissolving in water 2 + h₂o ⇔ hpo₄⁻² + h₃o⁺¹ iii. then write the expression for the kₐ kₐ = hpo₄⁻²h₃o⁺¹ / 3 iv. then solve for h₃o⁺¹ and input the numbers h₃o⁺¹ = (kₐ 4) / hpo₄⁻² = (6.20 × 10⁻⁸)(5) / (5.55 × 10⁻⁵) = 6

Explanation:

Step1: Identify the acid

The acid is \( \ce{H2PO4^{-1}} \) because it can donate a proton (has more acidic hydrogens compared to \( \ce{HPO4^{-2}} \)).

Step2: Write the acid dissociation reaction

The reaction is \( \ce{H2PO4^{-1}} + \ce{H2O}
ightleftharpoons \ce{HPO4^{-2}} + \ce{H3O^{+1}} \), so the blank (2) is \( \ce{H2PO4^{-1}} \).

Step3: Write the \( K_a \) expression

For the reaction \( \ce{H2PO4^{-1}} + \ce{H2O}
ightleftharpoons \ce{HPO4^{-2}} + \ce{H3O^{+1}} \), the \( K_a \) expression is \( K_a = \frac{[\ce{HPO4^{-2}}][\ce{H3O^{+1}}]}{[\ce{H2PO4^{-1}}]} \), so blank (3) is \( \ce{H2PO4^{-1}} \).

Step4: Solve for \( [\ce{H3O^{+1}}] \)

Rearranging the \( K_a \) expression: \( [\ce{H3O^{+1}}] = \frac{K_a [\ce{H2PO4^{-1}}]}{[\ce{HPO4^{-2}}]} \). So blank (4) is \( [\ce{H2PO4^{-1}}] \), blank (5) is \( 4.44 \times 10^{-5} \) (the concentration of \( \ce{H2PO4^{-1}} \)).
Now calculate: \( [\ce{H3O^{+1}}] = \frac{(6.20 \times 10^{-8})(4.44 \times 10^{-5})}{5.55 \times 10^{-5}} \)
First, multiply the numerator: \( (6.20 \times 10^{-8})(4.44 \times 10^{-5}) = 6.20 \times 4.44 \times 10^{-13} \approx 27.528 \times 10^{-13} = 2.7528 \times 10^{-12} \)
Then divide by \( 5.55 \times 10^{-5} \): \( \frac{2.7528 \times 10^{-12}}{5.55 \times 10^{-5}} = \frac{2.7528}{5.55} \times 10^{-7} \approx 0.496 \times 10^{-7} = 4.96 \times 10^{-8} \) (blank 6).

Answer:

1. \( \ce{H2PO4^{-1}} \)
2. \( \ce{H2PO4^{-1}} \)
3. \( \ce{H2PO4^{-1}} \)
4. \( [\ce{H2PO4^{-1}}] \)
5. \( 4.44 \times 10^{-5} \)
6. \( 4.96 \times 10^{-8} \, \text{M} \) (approximate value, depending on calculation precision)