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chemistry pH question HELP

(e) An acidic buffer solution is obtained when sodium ethanoate is dissolved in aqueous ethanoic acid.
(i) Calculate the pH of the buffer solution formed at 298 K when 0.125 mol of sodium ethanoate is dissolved in 250 cm3 of a 1.00 mol dm–3 solution of ethanoic acid.
The acid dissociation constant, Ka, for ethanoic acid is 1.70 × 10–5 mol dm–3 at 298 K.

the mark scheme:
Ka = [H+] [A–] / [HA] (1)
= [H+] [0.125 × 4] (1) / 1.00
[H+] = 1.70 × 10–5 / 0.125 × 4 = 3.40 × 10–5 (1)
pH = –log10 [H+] = 4.47 (1)

can someone explain why you multiply 0.125 by 4 ?

OP

help pls

Original post by 2016KK

help pls

Moles of ethanoic acid:

\frac{250}{1000} \times 1 = \frac{1}{4}

K_a = \frac{[H^+] [A^-]}{[HA]} = \frac{[H^+] \times 0.125}{(\frac{1}{4})} = [H^+] \times \frac{0.125}{(\frac{1}{4})} = [H^+] \times (0.125 \times 4)

(edited 6 years ago)

Original post by Swissblade

Moles of ethanoic acid:

\frac{250}{1000} \times 1 = \frac{1}{4}

K_a = \frac{[H^+] [A^-]}{[HA]} = \frac{[H^+] \times 0.125}{(\frac{1}{4})} = [H^+] \times \frac{0.125}{(\frac{1}{4})} = [H^+] \times (0.125 \times 4)

Yes but why

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