1. Calculate the pH at the equivalence point in titrating 0.100 M solutions of e
ID: 880017 • Letter: 1
Question
1. Calculate the pH at the equivalence point in titrating 0.100 M solutions of each of the following with 8.0×102MNaOH.
a). chlorous acid (HClO2)
b). benzoic acid (C6H5COOH)
2. How many milliliters of 0.105 MHCl are needed to titrate each of the following solutions to the equivalence point?
a). 40.0 mL of 9.50×102 MNaOH
b). 23.0 mL of 0.118 MNH3
c). 125 mL of a solution that contains 1.30 g of NaOH per liter
3. Calculate the pH of a solution that is made by combining 55 mL of 6.0×102 M hydrofluoric acid with 125 mL of 0.10 Msodium fluoride
Explanation / Answer
Calculate the pH at the equivalence point in titrating 0.100 M solutions of each of the following with 8.0×102MNaOH.
a). chlorous acid (HClO2)
Solution :- ka of the HClO2 = 1.1E-2
If we assume volume of the acid used = 10 ml then volume of NaOH needed
= 0.100 M * 10 ml / 0.08 M
=12.5 ml
So after the reaction all of acid will convert into its conjugate base
Thereof we need the kb value of the ClO2^-
Kb = kw / ka
= 1E-14 /1.1E-2
= 1E-12
After mixing the solutions volume will be increased to 10 ml + 12.5 ml = 22.5 ml = 0.0225 L
Therefore need to calculate the new molarity of the HClO2 = 0.100 M * 10 ml / 22.5 ml = 0.0444 M
Therefore moles of OCl2^- produced after reaction = 0.0444 M
Now lets write the kb equation for the OCl-
OCl- + H2O ---- > HClO2 + OH-
Kb = [HClO2][OH-]/[OCl-]
1E-12 = [x][x]/[0.0444]
1E-12 * 0.0444 = x^2
4.44E-14 =x^2
Taking square root of both sides we get
2.11E-7 =x
Now lets calculate the pOH using this concentration of OH-
pOH = -log[OH-]
pOH = - log [2.11E-7]
pOH = 6.7
pH = 14 – pOH
pH= 7.3
b). benzoic acid (C6H5COOH)
ka of benzoic acid = 6.3E-5
lets find the kb
kb = 1E-14 / 6.3E-5
kb = 1.6E-10
now lets make the kb equation
volume and concentrations will be same
after reaction all of the benzoic acid will converted to benzoate anion which will acts as base
kb = [C6H5COOH][OH-] / [C6H5COO-]
1.6E-10 = [x][x]/[0.0444]
1.6E-10 * 0.0444 = x^2
7.1E-12 =x^2
Taking square root of both sides we get
2.7E-7 = x =[OH-]
Now lets find pOH
pOH = -log[OH-]
= - log [2.7E-6]
=5.57
pH = 14 – pOH
pH = 14 – 5.57
pH = 8.43
How many milliliters of 0.105 MHCl are needed to titrate each of the following solutions to the equivalence point?
a). 40.0 mL of 9.50×102M NaOH
solution :- volume of HCl = 40 ml * 9.50×102M NaOH / 0.105 M HCl = 36.2 ml HCl
b). 23.0 mL of 0.118 MNH3
solution :- 23.0 ml * 0.118 M NH3 / 0.105 M HCl = 25.8 ml HCl
c). 125 mL of a solution that contains 1.30 g of NaOH per liter
Solution :- 1.30 g NaOH * 1 mol / 40.0 g = 0.0325 mol NaOH
Therefore moles of HCl = 0.0325 because mole ratio is 1 : 1
Volume of HCl = moles / molariy
=0.0325 mol / 0.105 mol per L
= 0.309 L
0.309 L * 1000 ml / 1 L = 309 ml HCl
Calculate the pH of a solution that is made by combining 55 mL of 6.0×102 M hydrofluoric acid with 125 mL of 0.10 Msodium fluoride
Solution :-
Lets first calculate the new molarities of the HF and NaF after mixing the solutions
Total volume after missing = 55 ml + 125 ml = 180 ml
New molarity of HF = 6.0×102 M * 55 ml / 180 ml =0.01833 M
New molarity of the NaF = 0.10 M * 125 ml / 180 ml =0.06944 M
HF is weak acid and F- is the conjugate base therefore this will form buffer solution so to calculate the pH need to use the Henderson equation.
pH= pka + log([base]/[acid])
pka of HF = 3.17
lets put the values in the formula
pH = 3.17 + log [0.06944/0.01833]
pH=3.17+0.58
pH= 3.75
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