Question

What molar ratio of sodium acetate to acetic acid should be used to prepare a buffer with pH = 4.5? `K_a` acetic acid = `1.8 x 10^-5`?

Answer 1

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

Explanation:

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

And thus,

`log_10((["AcO"^-])/(["HOAc"]))=pK_a-pH=-log_10(1.8xx10^-5)-4.5=4.75-4.5=0.245`

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