Calculate the pH for each of the following cases in the titration of 50.0mL of 0

Question

Calculate the pH for each of the following cases in the titration of 50.0mL of 0.130M HClO(Aq) with 0.130M KOH(Aq). Ionization constant for HClO is 4.8x10^-8
A) before any KOH is added B) after the addition of 25.0mL of KOH C) after addition of 35.0mL of KOH D) after the addition of 50.0 mL of KOH E) after the addition of 60.0mL of KOH
Calculate the pH for each of the following cases in the titration of 50.0mL of 0.130M HClO(Aq) with 0.130M KOH(Aq). Ionization constant for HClO is 4.8x10^-8
A) before any KOH is added B) after the addition of 25.0mL of KOH C) after addition of 35.0mL of KOH D) after the addition of 50.0 mL of KOH E) after the addition of 60.0mL of KOH
Ionization constant for HClO is 4.8x10^-8
A) before any KOH is added B) after the addition of 25.0mL of KOH C) after addition of 35.0mL of KOH D) after the addition of 50.0 mL of KOH E) after the addition of 60.0mL of KOH

Explanation / Answer

A:

HClO ----> H+ + ClO-

0.130M       0       0    begginig

0.130-x       x        x    equilibrium

K= 4.8 x 10-8= [ClO-][H+]/[HClO]= x2/(0.130-x) ------> x= 7.9 x 10-5 M= [H+] ---> pH= -log[H+]--->pH=4.1

B:

HClO + KOH ----> H2O + KClO

In the next four cases we have to find the reagent that is in excess, if there is one.

mol KOH= 0.130M x 0.025 L = 3.25 x 10-3mol

1 mol HClO ----- 1 mol KOH

        x         ------ 3.25 x 10-3mol KOH

x=3.25 x 10-3mol HClO

Now we need mol of HClO in the 50 mL solution:

mol HClO= 0.130M x 0.050L= 6.5 x 10-3mol ----> we only need 3.25 x 10-3mol HClO to react with 25 mL of KOH so, the HClO is in excess. The real reaction is:

HClO + KOH ----> H2O + KClO + HClO

This HClO will dissociate: HClO ----> H+ + ClO-

We need to find the concentration of HClO in excess:

Molarity= mol/volume =(6.5 x 10-3mol - 3.25 x 10-3mol)/ (0.050L + 0.025L)= 0.043M

HClO ----> H+ + ClO-

0.043         0        0

0.043-x     x         x

K= 4.8 x 10-8= [ClO-][H+]/[HClO]= x2/(0.043-x) ----> x= 4.54 x 10-5M=[H+] --->pH= -log[H+]--->pH= 4.3

C: you have to do exactly the same just change the volume:

mol KOH= 0.130M x 0.035 L = 4.55 x 10-3mol

1 mol HClO ----- 1 mol KOH

x ------ 4.55 x 10-3mol KOH

x=4.55 x 10-3mol HClO

Now we need mol of HClO in the 50 mL solution:

mol HClO= 0.130M x 0.050L= 6.5 x 10-3mol ----> we only need 4.55 x 10-3mol HClO to react with 25 mL of KOH so, the HClO is in excess. The real reaction is:

HClO + KOH ----> H2O + KClO + HClO

This HClO will dissociate: HClO ----> H+ + ClO-

We need to find the concentration of HClO in excess:

Molarity= mol/volume =(6.5 x 10-3mol - 4.55 x 10-3mol)/ (0.050L + 0.035L)= 0.023M

HClO ----> H+ + ClO-

0.023       0          0

0.023-x      x         x

K= 4.8 x 10-8= [ClO-][H+]/[HClO]= x2/(0.023-x) ----> x= 3.32 x 10-5M=[H+] --->pH= -log[H+]--->pH= 4.5

D:

mol KOH= 0.130M x 0.05 L = 6.5 x 10-3mol

1 mol HClO ----- 1 mol KOH

x ------ 6.5 x 10-3mol KOH

x=6.5 x 10-3mol HClO

Now we need mol of HClO in the 50 mL solution:

mol HClO= 0.130M x 0.050L= 6.5 x 10-3mol ----> in this case all the HClO is neutralized by the KOH because we have the same amount of mol for KOH and HClO. So, the pH= 7.

E:

mol KOH= 0.130M x 0.06 L = 7.8 x 10-3mol

1 mol HClO ----- 1 mol KOH

x ------ 7.8 x 10-3mol KOH

x=7.8 x 10-3mol HClO

Now we need mol of HClO in the 50 mL solution:

mol HClO= 0.130M x 0.050L= 6.5 x 10-3mol ----> in this case the KOH is in excess because we need 6.5 x 10-3mol of KOH to react with all the HClO. The real reaction is:

HClO + KOH ----> H2O + KClO + KOH

This KOH will dissociate: KOH ----> K+ + OH-

We need to find the concentration of KOH in excess:

Molarity= mol/volume =(7.8 x 10-3mol - 6.5 x 10-3mol)/ (0.050L + 0.060L)= 0.0118M

KOH is a strong base so:

[KOH]=[K+]=[OH-] -----> pOH= -log [OH-]= -log 0.0118= 1.9 ---> pH + pOH=14 ---> pH= 12.1

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