If the ?aof a monoprotic weak acid is 7.5×10−6, what is the pH of a 0.21 Msolution of this acid?
PLEASE BREAK IT DOWN STEP BY STEP
Ka=7.5×10-6(given)
Concentration=0.21M.Assume HA is the weak monoprotic acid.
HA<==> H+ + A- . Ka =
7.5×10-6
0.21. 0 ......0 . initial
-x .........+x ... +x ................. change
0.21-x ... x ..... x .................. equilibrium
Ka = [H+][A-] / [HA] =(x×x)/(0.21-x)
7.5x10-6= x² / (0.21-x)
As x is very small than 0.21.we can take 0.21-x as 0.21.
X2=0.21×7.5×10-6=1.575×10-6
X=√1.575×10-6=1.2×10-3
Therefore x=[H+]=1.2×10-3
pH=-log[H+]=-log(1.2×10-3)=2.9
pH is 2.9
Thank you
If the ?aof a monoprotic weak acid is 7.5×10−6, what is the pH of a 0.21...
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