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A chemist titrates 103.0 mL of a 0.4700 M solution of acetic acid (HCH ₃ CO ₂) with a 1.200 M solution of KOH. The pKa of acetic acid is 4.70. Calculate the pH of the acid solution after the chemist has added 9.891 mL of the KOH solution to it.

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Your first reaction might be a little something like this 

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Your first reaction might be a little something like this 

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But do not panic friends, just breathe and let your brains do the work

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Write the balanced chemical equation of the solution: 

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Write the balanced chemical equation of the solution: 

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KOH + HCH₃CO₂ --> H₂O + KCH₃CO₂

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Next, find the volume at equivalence using M1V1=M2V2:

M1 = 0.4700 M            M2 = 1.200 M
V1 = 103.0 mL            V2 = ?

We don't know what V2 is so we have to find by rearranging the M1V1 = M2V2 formula. We must rearrange it into V2 = M1V1 / M2

Now we just plug in our previous values, so it should look like this:

(0.4700 M)(103.0 mL) / (1.200 M) = 40.34 mL KOH

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Because we got 40.34 mL of KOH, that means that we are before equivalence because in the problem it states the chemist only added 9.891 mL of KOH.

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Because we got 40.34 mL of KOH, that means that we are before equivalence because in the problem it states the chemist only added 9.891 mL of KOH.

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So to find the pH before equilibrium, we will
1) write the equation for the weak substance which in this case would be HCH3CO2:

OH⁻ + HCH ₃ CO ² --> H+ + CH ₃ CO ₂-

2) Identify the moles of your weak acid and strong base
HCH3CO2 (weak acid) : 0.04034 mol  *the weak acid is in equilibrium*

KOH (strong base) : 0.009891 mol

3) Now for the fun part, we make a BCA table. This looks a lot like an ICE but instead it stands for Before-Change-After. It will look something like this:

OH⁻ +       HCH ₃ CO ² -->  H+   +  CH ₃ CO ₂-
B  0.009891    0.04034            -            0

C -0.009891   -0.009891        -     +0.009891

A        0              0.0304             -      0.009891

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In the chemical equation, the value of H+ will always be 0 (as a wise professor once said, if it doesn't pay your bills, don't pay attention to it.)

Always divide by the smallest number in the C row because you want the value to be 0 in the after row

CH3CO2- will be 0 because it was not added in the titration

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Now we will use the Henderson-Hasselbach equation to find pH:

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Now we will use the Henderson-Hasselbach equation to find pH:

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pH = pKa + log ([A-]/[HA])

4.70 + log (0.009891/0.0304) = 4.21

pH = 4.21 :)

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Now lets try to calculate pH after equivalence

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Now lets try to calculate pH after equivalence

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Find the relationship between Ka and pKa

10^-4.70 = 1.99 x 10^-5

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Next, use the equation to make the ICE table: 

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Next, use the equation to make the ICE table: 

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*Similarly like in the other table, H20 will not take part in the reaction. It doesn't pay your bills

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Substitute the equilibrium expressions from your ICE table into the Kb equation for HCH3CO2:  

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Substitute the equilibrium expressions from your ICE table into the Kb equation for HCH3CO2:  

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x2 / 0.4700 = 5.07 x 10^-10

solve for x:
x = (0.4700)(5.07 x 10^-10)
*the value you get from multiplying these two is square rooted*

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To calculate H3O+, use the ion product of water which is 1.01 x 10^-14:

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To calculate H3O+, use the ion product of water which is 1.01 x 10^-14:

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[H₃O₊] = Kw/[OH-] = (1.01 x 10^-14) / (1.543 x 10^-10)

= 6.545 x 10^-10

To find the pH = input this value into the -log:

pH = -log [H₃O₊] = -log (6.545 x 10^-10) = 9.18

This means pH = 9.18

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Finally, we will find the pH at equivalence: 

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Finally, we will find the pH at equivalence: 

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1) write the equation:

CH3CO2- + H2O ---> HCH3CO2 + OH-

*each mole of HCH3CO2 produces 1 mol of CH3CO2-

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= 0.04841
Then divide by 1.200 M:

0.04841  / 1.200M = 0.04034 L

Find the total volume of the solution:

0.103 mL + 0.04043 = 0.14334 L

You always convert your mL to Liters which is why it is now 0.103 mL

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Divide moles of CH3CO2 by total volume: 

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Divide moles of CH3CO2 by total volume: 

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0.04841/0.14334 = 0.337 M

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Plug the expression from the reaction into the Kb equation for CH3CO2- : 

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Plug the expression from the reaction into the Kb equation for CH3CO2- : 

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x = (0.337)(1.01 x 10^-14) / 10^-4.70 *the whole equation is square rooted

It can be kinda complicated to input this into a calculator but my best recommendation is to input the square root and within the square root input (0.337 x (1.01x10^-14) / 10^-4.70)

The answer will be
x = 1.31 x 10^-5

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Calculate H3O+ using the ion product of water: 

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Calculate H3O+ using the ion product of water: 

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[H3O+] = Kw/[OH-] = (1.01 x10^-14) / (1.31 x 10^-5)

= 7.73 x 10^-10

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Find the pH using the definition of pH: 

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Find the pH using the definition of pH: 

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pH = -log [H3O+] = -log (7.73 x 10^-10) = 9.11

pH = 9.11

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See, its not so bad. You did great! Bill Nye the science guy is happy for you :)

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See, its not so bad. You did great! Bill Nye the science guy is happy for you :)