What molar ratio of sodium acetate to acetic acid should be used to prepare a buffer with pH = 4.5? K_aKa acetic acid = 1.8 x 10^-51.8x105?

1 Answer
Jan 10, 2017

(["AcO"^-])/(["HOAc"])=1.76[AcO][HOAc]=1.76

Explanation:

pH=pK_a-log_10((["AcO"^-])/(["HOAc"]))pH=pKalog10([AcO][HOAc])

And thus,

log_10((["AcO"^-])/(["HOAc"]))=pK_a-pH=-log_10(1.8xx10^-5)-4.5=4.75-4.5=0.245log10([AcO][HOAc])=pKapH=log10(1.8×105)4.5=4.754.5=0.245

And thus 10^0.245=([AcO^-])/([HOAc])=1.76100.245=[AcO][HOAc]=1.76