Question

I did a lab and it's asking for the theoretical pH with the addition of HCl. The procedure:measure out 5.0ml of 1.0ml sodium acetate solution in a 100 ml graduated cylinder. dilute the 5.0 ml w/ water to a final volume 50.0 mL&stir.measure out 1.00ml of 1.0 M HCl&add it to the sodium acetate solution.stir the mixture.calculate the theoretical pH.

Answer 1

Lets convert the given concentrations to moles

5ml of 1M (typo corrected) CH₃COONa : is equal to

number of moles = molarity * volume in liter

= 1 * 0.005 lit = 0.005 moles

Similarly 1.0 mL of 1 .0 M HCl is equal to 0.001 moles

The reaction that takes place

CH₃COONa + HCl --> CH₃COOH + NaCl

According to this equation in final solution,

Amount of HCl = 0.001 mol -0.005 mol = 0.005 mol

Amount of Acetic acid formed = 0.005 mol

Strong acid (HCl) are fully ionised but weak acid (Acetic acid) are only partly ionised in solution.

In mixture, HCl + CH3COOH H⁺+ Cl ^- + H⁺ + CH₃COO^-.

The strong dissociation of HCl drives the H⁺ and   CH₃COO^- to the left of the equilibrium. So, we can neglect weak acid contribution to pH.

In final solution 0.005 moles of HCl is present in 51 mL of solution.

Concentration of HCl = Cocentration of H⁺ = (0.005/ 51)*1000 = 0.098 M

pH = -log [H+] = -log 0.098 = -1.0.