Question

A buffer is prepared by adding 20.0 g of sodium acetate 1ch3coona2 to 500 ml of a 0.150 m acetic acid 1ch3cooh2 solution. (a) determine the ph of the buffer. (b) write the complete ionic equation for the reaction that occurs when a few drops of hydrochloric acid are added to the

Answer 1

a) m(CH₃COONa) = 20 g.

n(CH₃COONa) = m(CH₃COONa) ÷ M(CH₃COONa).

n(CH₃COONa) = 20 g ÷ 82.034 g/mol.

n(CH₃COONa) = 0.244 mol; amount of substance.

V(CH₃COONa) = 500 ml ÷ 1000 ml/L.

V(CH₃COONa) = 0.5 L.

c(CH₃COONa) = n(CH₃COONa) ÷ V(CH₃COONa).

c(CH₃COONa) = 0.244 mol ÷ 0.5 L.

c(CH₃COONa) = 0.488 M; molarity of sodium acetate.

c(CH₃COOH) = 0.150 M; molarity of acetic acid.

Ka(CH₃COOH) = 1,8·10⁻⁵.

pKa = -logKa = 4,75.

Henderson–Hasselbalch equation: pH = pKa + log(cs/ck).

pH = 4.75 + log(0.488M / 0.150M).

pH = 5.262; pH of buffer solution.

b) Chemical reactions:

1) HCl(aq) → H⁺(aq) + Cl⁻(aq).

2) CH₃COONa(aq) → CH₃COO⁻(aq) + Na⁺(aq).

3) CH₃COO⁻(aq) + H⁺(aq) ⇄ CH₃COOH.

Sum: CH₃COONa + HCl ⇄ CH₃COOH + NaCl.

Answer 2

first, we have to get moles of CH3COONa = mass/molar mass

= 20 g / 82.03 g/mol = 0.244 moles

when we have [CH3COOH] = 0.15 M

∴ [CH3COONa] = moles of CH3COONa / Volume of solution

= 0.244 moles / 0.5L = 0.488 M


when we look up for Ka of acetic acid value it is equal 1.8 x 10^-5

So we can get Pka = -㏒Ka

= -㏒(1.8 x10^-5)
= 4.7

now we will use Henderson - Hasselbalchn equation to get the PH:

PH = Pka + ㏒[conjugate basic/weak acid]

when CH3COOH is the weak acid & CH3COO- is the conjugate base so by substitution:

PH = 4.7 + ㏒ (0.488/0.15)

= 5.2

b) when we have this equation for the reaction:

HCl + CH3COONa → CH3COOH + NaCl

ionic equation : H+ + Cl- + CH3COO- + Na+ → CH3COOH + Na+ + Cl-

when HCl + H2O → H3O+ + Cl-

∴ the reaction will be:

CH3COO- (aq) + H3O+(aq) → CH3COOH(aq) + H2O(l)