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#?

1 Answer
anor277
Jan 10, 2017

#(["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#

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