calculate the ph of a solution prepared by dissolving 0.150 mol of benzoic acid and 0.300 mol of sodium benzoate in water sufficient to yield.1.00l of solution
Answers
Answer:
Answer : The pH of a solution is, 4.50
Explanation : Given,
K_a=6.30\times 10^{-5}K
a
=6.30×10
−5
Concentration of benzoic acid (Acid) = 0.150 M
Concentration of sodium benzoate (salt) = 0.300 M
First we have to calculate the value of pK_apK
a
.
The expression used for the calculation of pK_apK
a
is,
pK_a=-\log (K_a)pK
a
=−log(K
a
)
Now put the value of K_aK
a
in this expression, we get:
pK_a=-\log (6.30\times 10^{-5})pK
a
=−log(6.30×10
−5
)
pK_a=5-\log (6.30)pK
a
=5−log(6.30)
pK_a=4.20pK
a
=4.20
Now we have to calculate the pH of buffer.
Using Henderson Hesselbach equation :
pH=pK_a+\log \frac{[Salt]}{[Acid]}pH=pK
a
+log
[Acid]
[Salt]
Now put all the given values in this expression, we get:
pH=4.20+\log (\frac{0.300}{0.150})pH=4.20+log(
0.150
0.300
)
pH=4.50pH=4.50
Thus, the pH of a solution is, 4.50
Hope it helps ☺️!!!
please mark the answer as brain liest
follow me to get the answers