Calculating the buffer solution

Calculating the pH of Buffered Solutions

A buffer solution is one in which the pH of the solution is "resistant" to small additions of either a strong acid or strong base. Buffers usually consist of a weak acid and its conjugate base, in relatively equal and "large" quantities. Calculations are based on the equation for the ionization of the weak acid in water forming the hydronium ion and the conjugate base of the acid. "HA" represents any weak acid and "A-" represents the conjugate base. A buffer system can be made by mixing a soluble compound that contains the conjugate base with a solution of the acid such as sodium acetate with acetic acid or ammonia with ammonium chloride. The above equation for Ka can be rearranged to solve for the hydronium ion concentration. By knowing the Ka of the acid, the amount of acid, and the amount of conjugate base, the pH of the buffer system can be calculated.

formula of calculating the buffer solution:

 

 

 

FOR examples:

1. 30 mL of 0.10M NaOH neutralised 25.0mL of hydrochloric acid. Determine the concentration of the acid!

Answered

 there are 6 steps:

1.Write the balanced chemical equation for the reaction
NaOH(aq) + HCl(aq) -----> NaCl(aq) + H₂O(l)

2.Extract the relevant information from the question:
NaOH V = 30mL , M = 0.10M HCl V = 25.0mL, M = ?

3.Check the data for consistency
NaOH V = 30 x 10^-3L , M = 0.10M HCl V = 25.0 x 10^-3L, M = ?          

4.Calculate moles NaOH
n(NaOH) = M x V = 0.10 x 30 x 10^-3 = 3 x 10^-3 moles

5.From the balanced chemical equation find the mole ratio
NaOH:HCl
1:1

6.Find moles HCl
NaOH: HCl is 1:1

So n(NaOH) = n(HCl) = 3 x 10^-3 moles at the equivalence point

Calculate concentration of HCl: M = n ÷ V

n = 3 x 10^-3 mol, V = 25.0 x 10^-3L

M(HCl) = 3 x 10^-3 ÷ 25.0 x 10^-3 = 0.12M or 0.12 mol L^-1

 

2. 30 mL of 0.10M NaOH neutralised 25.0mL of hydrochloric acid. Determine the concentration of the acid!

Answered

 There are 7 steps

1.Write the balanced chemical equation for the reaction
NaOH(aq) + H₂SO₄(aq) -----> Na₂SO₄(aq) + 2H₂O(l)

2.Extract the relevant information from the question:
NaOH V = 50mL, M = 0.2M H₂SO₄ V = 20mL, M = ?

3.Check the data for consistency
NaOH V = 50 x 10^-3L, M = 0.2M H₂SO₄ V = 20 x 10^-3L, M = ?

4.Calculate moles NaOH
n(NaOH) = M x V = 0.2 x 50 x 10^-3 = 0.01 mol

5.From the balanced chemical equation find the mole ratio
NaOH:H₂SO₄
2:1

6.Find moles H₂SO₄
NaOH: H₂SO₄ is 2:1

So n(H₂SO₄) = ½ x n(NaOH) = ½ x 0.01 = 5 x 10^-3 moles H₂SO₄ at the

equivalence point

7.Calculate concentration of H₂SO₄: M = n ÷ V
n = 5 x 10^-3 mol, V = 20 x 10^-3L

M(H₂SO₄) = 5 x 10^-3 ÷ 20 x 10^-3 = 0.25M or 0.25 mol L^-1

to be continued..,!

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