Question

An analytical chemist is titrating 179.5ml of 0.8600Ma solution of cyanic acid(HCNO) with0.6500m a solution of NaOH. The PKa of cyanic acid is 3.46 . Calculate the pH of the acid solution after the chemist has added 274.6ml of NaOH the solution to it.

Answer 1

12.73

Step-by-step explanation:

Write the balanced equation first:

`HCNO (aq) + NaOH (aq)\rightarrow NaCNO (aq) + H_2O (l)`

Find moles of the acid (subscript of 1 for simplicity):

`n_1 = 0.8600 M\cdot 0.1795 L = 0.15437 mol`

The number of moles of NaOH at the equivalence point will be the same, as we have a 1 : 1 stoichiometry (use subscript of 2 for NaOH):

`n_2 = n_1 = 0.15437 mol`

Find the equivalence volume of NaOH:

`V_(eq) = (n_2)/(c_2) = (0.15437 mol)/(0.6500 M) = 237.5 mL`

Since we're looking at an instant where we actually add a total of 274.6 mL, this is much greater than the equivalence volume, so we'll have a huge excess of NaOH and we can ignore any ionization of the acid, as it'll be negligible compared to the hydroxide concentration.

Find the excess volume of NaOH added past the equivalence point:

`V_2 = 274.6 mL - 237.5 mL = 37.1 mL`

Find the molarity of excess NaOH using the dilution formula (we're dividing by a total volume of this solution):

`c_2 = (0.6500 M\cdot 37.1 mL)/(179.5 mL + 274.6 mL) = 0.053105 M`

Find the pOH value:

`pOH = -log[NaOH] = -log(0.053105) = 1.27`

Then:

`pH = pK_w - pOH = 14.00 - 1.27 = 12.73`