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A 0.25 mol sample of a weak acid with an unknown pKa was combined with 10...

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?

Homework Answers

Answer #1

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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