A 0.25 mol sample of a weak acid with an unknown pKa was combined with 10 mL of 3.00 M KOH, and the resulting solution was diluted to 1.500 L. The measured pH of the solution was 3.85. What is the concentration of all species in the solution? What will the pH of the solution be if an additional 10.00 mL of KOH is added?
moles of KOH in 10 ml of its solution=0.03
because 1000 ml include 3 moles.
so 1 ml include 3/1000 moles
10 ml include (3/1000)*10 = 0.03 moles
1.5L+10/1000L is approximatly equal to 1.5L
now after dilution conc. of KOH is =0.03/1.5 = 0.02M
conc. of K+ =0.02M
KOH ---------> K+ OH-
conc. of weak acid C = 0.25/1.5 = 0.166M
Let weak acid be HX
HX --------> H+ X-
C-C@ C@ C@
As we know H+ and OH- reacts to form H2O so conc. of H+ decrease
Net conc. of H+ ions is P = C@ - 0.02
-log(P) =3.85
antilog(logP) = antilog (-3.85)
P =0.000141
C@ - 0.02 =0.000141
C@ = 0.020141
conc. of X- = 0.020141M
conc. of H+ = 0.000141M
if 10 ml KOH is added conc of KOH is 2 * 0.02 =0.04M
but equilibirium of weak acid now shift in forward direction because OH- conc increase
so there is slight tends to zero change in ph because forward shift of equilibirium compansate it
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