Chemistry, 25.07.2019 16:00 Tyrant4life

For the titration of 50.0 ml of 0.020 m aqueous salicylic acid with 0.020 m koh (aq), calculate the ph after the addition of 55.0 ml of the base. for salycylic acid, pka = 2.97.'

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For the titration of 50.0 mL of 0.020 M aqueous salicylic acid with 0.020 M KOH, the pH of the solution after the addition of 55.0 mL of KOH is 10.98.

The reaction between salicylic acid and KOH is:

C₇H₆O₃(aq) + KOH(aq) ⇄ C₇H₅O₃⁻K⁺(aq) + H₂O(l)

The number of moles of salicylic acid and KOH before the reaction is:

$n_{{C_{7}H_{6}O_{3}}_{i}} = 0.020 mol/L*0.050 L = 1.0 \cdot 10^{-3} \:mol$

$n_{{KOH}_{i}} = 0.020 mol/L*0.055 L = 1.1 \cdot 10^{-3} \:mol$

After the addition of KOH, all the acid will react with KOH, and the number of moles of KOH remaining in the solution will be:

$n_{{KOH}_{f}} = n_{{KOH}_{i}} - n_{{C_{7}H_{6}O_{3}}_{i}} = 1.1 \cdot 10^{-3} \:mol - 1.0 \cdot 10^{-3} \:mol = 1.0 \cdot 10^{-4} \:moles$

We need to find the concentration of this base with the total volume of solution:

$C_{KOH} = \frac{n_{{KOH}_{f}}}{V_{t}} = \frac{1.0 \cdot 10^{-4} \:moles}{0.050 L + 0.055 L} = 9.52 \cdot 10^{-4} mol/L$

Now, with this concentration we can find the pH as follows:

pH = 14 - pOH

$pH = 14 - (-log[OH^{-}])$

$pH = 14 + log(9.52 \cdot 10^{-4}) = 10.98$

Therefore, the pH of the solution is 10.98.

You can learn more about pH here:

linklink

I hope it helps you!

Answer from: fixianstewart

First by getting moles of salicylic acid:

moles of salicylic acid = molarity * volume

= 0.02 * 0.05 L

= 0.001 mol

then moles of base KOH = molarity * volume

= 0.02 * 0.055L

= 0.0011

when total volume = 0.05 + 0.055 = 0.105 L

[salisalic acid] = moles / total volume

= 0.001 / 0.105

= 0.0095

[KOH] = moles / total volume

= 0.0011 / 0.105

= 0.01

by using H-H equation, we can get the PH:

PH = Pka + ㏒[salt/acid]

by substitution:

PH = 2.97 + ㏒[0.01 / 0.0095]

= 2.99
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For the titration of 50.0 ml of 0.020 m aqueous salicylic acid with 0.020 m koh (aq), calculate the...

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