Question

Calculate the pH for each of the following cases in the titration of 50.0 mL of 0.210 M HClO(aq) with 0.210 M KOH(aq).

Answer 1

Answer:

(a) before addition of any KOH  = pH =-log [0.210]=0.68

(b) after addition of 25.0 mL of KOH  = pH =-log [0.210]=1.15

(c) after addition of 35.0 mL of KOH  = pH =-log [0.210]=1.43

(d) after addition of 50.0 mL of KOH  = pH =-log [0.210]=7.00

(e) after addition of 60.0 mL of KOH = pH =-log [0.210]=12.30

Explanation:

pH is calculated as follows:

pH = -log [H+], for acids

pOH = -log [OH-], for bases

pH = 14 - pOH

In this tritation, we add a base (KOH), to an acid (HClO), which neutralices each other in a stoiquiometric relationship of 1:1

So, reemplacing the values in each case we got:

(a) before addition of any KOH

: The concentration is equal to the concentration of HClO, beacuse no KOH has been added

pH =-log [0.210]=0.68

(b) after addition of 25.0 mL of KOH  

The concentration is equal to the amount of moles of HClO, that haven´t reacted with the moles of KOH cantained in the 25 ml added, divided by the total volumen (50 ml + 25 ml) = 75 ml

to find the amount of moles in the 25ml, and the 50 ml:

moles KOH = volumen L * Molarity = 0.025 ml * 0.210 M = 5.25*$10^{-3}$

moles HClO = volumen L * Molarity = 0.050 ml * 0.210 M = 1.05*$10^{-2}$

moles HClO - moles KOH = 1.05*$10^{-2}$-5.25*$10^{-3}$=5.25*$10^{-3}$

concentration = moles / volume = $\frac{5.25*10^{-3}moles of HClO}{0.075ml}$ = 0.070M

pH =-log [0.070]=1.15

(c) after addition of 35.0 mL of KOH  

The concentration is equal to the amount of moles of HClO, that haven´t reacted with the moles of KOH cantained in the 35 ml added, divided by the total volumen (50 ml + 35 ml) = 85 ml

to find the amount of moles in the 35ml, and the 50 ml:

moles KOH = volumen L * Molarity = 0.035 ml * 0.210 M = 7.35*$10^{-3}$

moles HClO = volumen L * Molarity = 0.050 ml * 0.210 M = 1.05*$10^{-2}$

moles HClO - moles KOH = 1.05*$10^{-2}$-7.35*$10^{-3}$=3.15*$10^{-3}$

concentration = moles / volume = $\frac{3.15*10^{-3}moles of HCl}{0.085ml}$ = 0.037M

pH =-log [0.037]=1.43

(d) after addition of 50.0 mL of KOH  

moles HCl = moles KOH

pH = 7

(e) after addition of 60.0 mL of KOH

The concentration is equal to the amount of moles of KOH, that haven´t reacted with the moles of HClO cantained in the initial 50 ml, because now the KOH is in excess, divided by the total volumen (50 ml + 60 ml) = 100 ml

to find the amount of moles in the 50ml, and the 60 ml:

moles KOH = volumen L * Molarity = 0.060 ml * 0.210 M = 1.26*$10^{-2}$

moles HClO = volumen L * Molarity = 0.050 ml * 0.210 M = 1.05*$10^{-2}$

moles KOH - moles HClO = 1.26*$10^{-2}$-1.05*$10^{-2}$=2.1*$10^{-3}$

concentration = moles / volume = $\frac{2.1*10^{-3}moles of KOH}{0.110ml}$ = 0.020M

pOH =-log [0.020]=1.70

pH = 14 - pOH = 12.3

Answer 2

35.0 mL of 0.210 M
KOH molarity = moles/volume
 find moles of OH do the same thing for: 50.0 mL of 0.210 M HClO(aq) but for H+ they will cancel out: H+ + OH- -> H2O
 but you'll have some left over,
 pH=-log[H+] pOH
     =-log[OH-] pH+pOH
     =14