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Calculating the pH at equi.. A chemist titrates 70.0 ml. of a 04557 M carbonic acid...


Calculating the pH at equi.. A chemist titrates 70.0 ml. of a 04557 M carbonic acid (H,co,) solution with o8352 M NaOlH Solut
Calculating the pH at equi.. A chemist titrates 70.0 ml. of a 04557 M carbonic acid (H,co,) solution with o8352 M NaOlH Solution at 2s c Calculate the pH at equivalence. The pk, of carbonic acid is 3.60 Round your answer to 2 decimal places. Note for advanced students: you may assume the total volume of the solution equals the initial volume plus the volume of NaoH Solution added. рH - D
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use:

pKa = -log Ka

3.6 = -log Ka

Ka = 2.512*10^-4

find the volume of NaOH used to reach equivalence point

M(H2CO3)*V(H2CO3) =M(NaOH)*V(NaOH)

0.4557 M *70.0 mL = 0.8352M *V(NaOH)

V(NaOH) = 38.1932 mL

Given:

M(H2CO3) = 0.4557 M

V(H2CO3) = 70 mL

M(NaOH) = 0.8352 M

V(NaOH) = 38.1932 mL

mol(H2CO3) = M(H2CO3) * V(H2CO3)

mol(H2CO3) = 0.4557 M * 70 mL = 31.899 mmol

mol(NaOH) = M(NaOH) * V(NaOH)

mol(NaOH) = 0.8352 M * 38.1932 mL = 31.899 mmol

We have:

mol(H2CO3) = 31.899 mmol

mol(NaOH) = 31.899 mmol

31.899 mmol of both will react to form HCO3- and H2O

HCO3- here is strong base

HCO3- formed = 31.899 mmol

Volume of Solution = 70 + 38.1932 = 108.1932 mL

Kb of HCO3- = Kw/Ka = 1*10^-14/2.512*10^-4 = 3.981*10^-11

concentration ofHCO3-,c = 31.899 mmol/108.1932 mL = 0.2948M

HCO3- dissociates as

HCO3- + H2O -----> H2CO3 + OH-

0.2948 0 0

0.2948-x x x

Kb = [H2CO3][OH-]/[HCO3-]

Kb = x*x/(c-x)

Assuming x can be ignored as compared to c

So, above expression becomes

Kb = x*x/(c)

so, x = sqrt (Kb*c)

x = sqrt ((3.981*10^-11)*0.2948) = 3.426*10^-6

since c is much greater than x, our assumption is correct

so, x = 3.426*10^-6 M

[OH-] = x = 3.426*10^-6 M

use:

pOH = -log [OH-]

= -log (3.426*10^-6)

= 5.4652

use:

PH = 14 - pOH

= 14 - 5.4652

= 8.5348

Answer: 8.53

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