Question

Calculate the expected pH of the solution at the equivalence point using YOUR AVERAGE VALUES for the concentrations of NaOH and acetic acid and the volumes of each that you used. (Just like in Prelab Q3, you will only have the conjugate base and spectator ions present at the equivalence point in a volume that is the sum of the volumes of acid, water, and base you combined.) The Ka for acetic acid is 1.8 x 10-5.'

Answer 1

Complete Question

The complete question is shown on the first uploaded image

Answer:

The pH is
`pH = 4.94`

Step-by-step explanation:

From the question we are told that

The average concentration of NaOH is
`[NaOH] = 0.101 M`

The volume of NaOH is
`V__(NaOH)} = 15.00 mL`

The average concentration of Acetic acid is
`[Acetic \ Acid] =0.497 \ M`

The volume of Acetic acid is
`V__(Acetic \ Acid)} = 5.00 \ mL`

The chemical equation for this reaction is

`NaOH + CH_3COOH ---> CH_3 COONa + H_2 O`

The total volume of the solution is

`V__(Total)} = V__(NaOH)} + V__(Acetic \ Acid)}`

Substituting values

`V__(Total)} = 15 + 5`

`V__(Total)} = 20mL = 20 *10^(-3) L`

The number of moles of NaOH is mathematically represented as

`n__(NaOH)} = [NaOH] * V__(NaOH)}`

substituting values

`n__(NaOH)} = 0.101 * 15*10^(-3)`

`n__(NaOH)} = 0.001515 \ moles`

The number of moles of Acetic acid is mathematically represented as

`n__(Acetic acid)} = [Acetic \ acid] * V__(Acetic acid)}`

substituting values

`n__(Acetic acid)} = 0.497 * 5*10^(-3)`

`n__(Acetic acid)} = 0.002485\ moles`

From the chemical equation

1 mole of NaOH reacts with 1 mole of Acetic acid to produce 1 mole of
`CH_3 COONa` salt and 1 mole of
`H_2 O`

So

0.001515 moles of NaOH reacts with 0.001515 moles of Acetic acid to produce 0.001515 moles of
`CH_3 COONa` salt and 0.001515 moles of
`H_2 O`

This implies the number of moles of NaOH remaining after the react would be

`\Delta n__(NaOH)} = 0.001515 - 0.001515`

`\Delta n__(NaOH)} = 0 \ mole`

the number of moles of Acetic acid remaining after the react would be

`\Delta n__(Acetic acid)} = 0.002485 - 0.001515`

`\Delta n__(Acetic acid)} = 0.00097 \ moles`

the number of moles of
`CH_3 COONa \ salt` remaining after the react would be

`\Delta n__(CH_3 COONa \ salt)} = 0 + 0.001515`

`\Delta n__(CH_3 COONa \ salt)} = 0.001515 \ moles`

the number of moles of
`H_2 O` remaining after the react would be

`\Delta n__(H_2O)} = 0 + 0.001515`

`\Delta n__(H_2O)} = 0.001515 \ moles`

The expected pH is mathematically evaluated as

`pH = pK_a + log [([CH_3 COONa])/([Acetic \ acid]) ]`

Where
`pKa` is mathematically evaluated as

`pK_a = - log (K_a)`

The concentration of
`CH_3 COONa \ salt` is mathematically evaluated a s

`[CH_3 COONa] = (\Delta n_CH_3 COONa \ salt )/(V__(Total))}`

substituting values

`[CH_3 COONa] = (0.001515)/(20 *10^(-3))`

`[CH_3 COONa] = 0.07575M`

The concentration of Acetic acid is mathematically evaluated as

`[Acetic acid] = (\Delta n__Acetic acid)/(V__(Total))}`

substituting values

`[CH_3 COONa] = (0.00097)/(20 *10^(-3))`

`[CH_3 COONa] = 0.0485 M`

Substituting values into the equation for pH

`pH = - log (1 .8 *10^(-5)) + log [(0.07575)/( 0.0485) ]`

`pH = log [(0.07575)/( 0.0485) ] - log (1 .8 *10^(-5))`

`pH = log [(1.561856)/(1.8*10^(-5)) ]`

`pH = 4.94`