Answer:
a) pH = 9.41
b) pH after adding 60.0 mL of 2.00 M HBr(aq) = 9.06
c) pH after adding 120.0 mL of 2.00 M HBr(aq) = 8.98
Explanation:
Given that:
mass of
mass of
Total volume = 750.0 mL = 0.75 L
Equation for the reaction
The molar mass for = 17 g/mol
molar mass of HCl = 36 g/mol
∴ number of moles =
For ;
number of moles =
= 0.3265 mole of
For HCl ;
number of moles =
= 0.1328 mole of HCl
The stoichiometric ratios for the reaction are in the ratio of 1:1
That implies that 1 mole of HCl reacts with 1 mole of
with HCl being the limiting reactant;
0.1328 mole of HCl react with 0.1328 mole of to form 0.1328 mole of
Final concentration of = 0.3265 - 0.1328
= 0.1937 mole of
Final concentration of HCl = 0.1328 -0.1328
= 0.00 mole of HCl
However, concentration of and can be calculated as follows:
Since Concentration(mol/L) =
For ; we have:
=
= 0.2583 mole/L
For ; we have :
=
= 0.1771 mole/L
Equation of the reaction in the solution of HBr is as follows:
Dissociation of yields:
Since, is present in both of the above reactions; then this a buffer solution which is demonstrated by using Henderson Hasslelbach equation.
given that Kb =
pKb = -log (Kb)
pKb =
pKb = 4.75
pH = 14 - pOH
pH = 14 - 4.5861
pH = 9.4139
pH = 9.41
b) After 2.00 M of HBr and Volume = 60.0 mL
Volume increases from 750mL to : (750 + 60 ) mL = 810 mL = 0.81 mL
Definitely the concentration of the Br⁻ will also increase.
The new concentration of [Br⁻] =
moles of Br⁻ added = The new concentration of [Br⁻] × volume
= 60 ×2.0
= 120 mL. mol
mole of Br⁻ already in the solution = 0.1771 +
= 0.1771 + 0.12
= 0.2971 mol of Br⁻
[Br⁻] =
[Br⁻] =
[Br⁻] = 0.3668 M
We need to determine the new concentration of again ; So using ; we have :
0.2583 × 750 = × 810
pH = 14 - pOH
pH = 14 - 4.9357
pH = 9.0643
pH = 9.06
c) After 2.00 M of HBr and Volume = 12.0 mL
Volume increases from 750mL to : (750 + 120 ) mL = 870 mL = 0.87 mL
Definitely the concentration of the Br⁻ will also increase.
The new concentration of [Br⁻] =
moles of Br⁻ added = The new concentration of [Br⁻] × volume
= 120 ×2.0
= 240 mL. mol
mole of Br⁻ already in the solution = 0.1771 +
= 0.1771 + 0.24
= 0.4171 mol of Br⁻
[Br⁻] =
[Br⁻] =
[Br⁻] = 0.480 M
We need to determine the new concentration of again ; So using ; we have :
0.2583 × 750 = × 870
pH = 14 - pOH
pH = 14 - 5.02
pH = 8.98