Question

Font 1.1. Paragraph 4.1 .5. 1. .2 . 1.1 ...6 Table 1. Raw Data Trial 1 Trial 2 unknown acid un acid mass of unknown acid transferred to beaker(@) 0.589 0.556 concentration of NaOH titrant solution (M) equivalence point volume of NaOH titrant (mL) 26.67 ml 26.16 ml half-equivalence point volume of NaOH titrant (mL) 13.84 ml 13.08 ml half-equivalence point pH 4.91 pH 5.14 pH 1. Calculate the moles of NaOH delivered at the equivalence point for each trial, then enter the values in Table 2. Show at least one sample calculation 2. Enter the number of moles of unknown acid as experimentally determined by the titration into Table 2. 3. Calculate the molar mass of the unknown acid for each trial, then enter the values in Table 2. Show at least one sample calculation

molariity of NaOH is 0.09591 M

Answer 1

1. Number of moles = Molarity * Volume in Litres

Trial 1 : moles of NaOH = 0.09591 M * 0.02667 L

moles of NaOH = 0.00255792 mol

Trial 2: moles of NaOH = 0.09591 M * 0.02616 L

= 0.0025090056 mol

2. As 1 mol base reacts with 1 mol acid,

Trial 1: number of moles of unknown acid = 0.00255792 mol

Trial 2: 0.0025090056 mol

3. number of moles = mass/molar mass

Trial 1: molar mass = mass/number of moles = 0.589 g/0.00255792 mol = 230.265 mol

Trial 2: 0.556 g/0.0025090056 mol = 221.602 g/mol

4. average molar mass = (230.265 + 221.602)/2 = 225.933 g/mol

5. p​​​​H at half equivalence point = p​​​​​​ka

Trial 1: pka = 4.91

Trial 2: pka = 5.14

6. Average pka= (4.91+5.14)/2 = 5.025