Answer :
The pH of the aqueous hydrocyanic acid solution after 36.9 mL of potassium hydroxide have been added is approximately 13.14.
The pH of an aqueous hydrocyanic acid solution titrated with a potassium hydroxide solution is calculated using stoichiometry. Given that a 20.4 mL sample of a 0.383 M aqueous hydrocyanic acid solution is titrated with a 0.318 M aqueous potassium hydroxide solution, the pH of the hydrocyanic acid solution after 36.9 mL of potassium hydroxide have been added is calculated as follows:Reaction Equation:H + OH- ⟶ H2OThe balanced chemical equation above shows that the reaction between hydrocyanic acid and potassium hydroxide is an acid-base reaction.
The first step in calculating the pH of an aqueous hydrocyanic acid solution titrated with a potassium hydroxide solution is to determine the number of moles of the solute (hydrocyanic acid) in the given volume of the solution, as shown below:Molarity (M) = moles of solute (n) / volume of solution (V in L) => n = MVIn the equation above, M is the molarity of the solution, n is the number of moles of solute, and V is the volume of the solution in liters.n = 0.383 M x 0.0204 L = 0.0078132 moles of hydrocyanic acidThe next step is to determine the number of moles of the titrant (potassium hydroxide) that have been added to the hydrocyanic acid solution.
Moles of potassium hydroxide added = molarity of potassium hydroxide x volume of potassium hydroxide addedn = 0.318 M x 0.0369 L = 0.0117502 moles of potassium hydroxideSince the reaction between hydrocyanic acid and potassium hydroxide occurs in a 1:1 ratio, it can be concluded that the number of moles of hydroxide ions (OH-) added is equal to the number of moles of hydrocyanic acid neutralized.Moles of hydroxide ions (OH-) added = 0.0078132 moles of hydrocyanic acid neutralized = 0.0078132 moles of OH-The total volume of the solution after 36.9 mL of potassium hydroxide have been added is 20.4 mL + 36.9 mL = 57.3 mL = 0.0573 L.
The concentration of OH- ions in the solution is given by:n(OH-) / V(solution) = 0.0078132 / 0.0573 = 0.1362559 MThe pOH of the solution is calculated as:pOH = -log[OH-] = -log(0.1362559) = 0.8641346The pH of the solution is obtained by subtracting the pOH from 14:pH = 14 - 0.8641346 = 13.1358654Therefore, the pH of the aqueous hydrocyanic acid solution after 36.9 mL of potassium hydroxide have been added is approximately 13.14.
Learn more about Hydrocyanic acid here,
https://brainly.com/question/31325872
#SPJ11