A buffer is made by combining 5.55 g of NH3(Kb= 1.8 X 10-5)with 4.78 g of HCl and diluting to a volume of 750.0 mL. a. What is the pH?b. What is the pH after adding 60.0 mL of 2.00 M HBr(aq)?c. What is the pH after adding 120.0 mL of 2.00 M HBr(aq)?

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

може ли да зареждате повече информация, моля