Calculate the pH of the solution resulting from the addition of 75.0 mL of 0.15 M KOH to 35.0 mL of 0.20 M HCN (Ka(HCN) = 4.9 × 10^–10)
millimols KOH initially = mL x M = 75.0 x 0.15M = 11.25
millimols HCN initially = 7.0
..........KOH + HCN ==> KCN + H2O
I...........0.........7.0.............0.........0
add......11.25..................................................
C........-7.0......-7.0..........7.0...........7.0
E..........4.25......0............7.0............7.0
From the E line you can see you have an excess of a very strong base (i.e., KOH) so the pH will be determined by the KOH. This is NOT a buffered solution.
pOH = -log(KOH)
Then pH + pOH = pKw = 14. You know pKw and pOH, solve for pH.
To calculate the pH of the resulting solution, we need to determine the concentration of the hydronium ion (H3O+) in the solution. This can be done using the principles of acid-base reactions and the equilibrium constant (Ka) for the weak acid, HCN.
Step 1: Write the chemical equation for the reaction between KOH and HCN:
KOH + HCN -> KCN + H2O
Step 2: Identify the limiting reagent:
To determine the limiting reagent, compare the number of moles of each reactant. The reactant with fewer moles will be completely consumed in the reaction, while the other reactant will be in excess.
In this case, we need to calculate the number of moles for each reactant:
Number of moles of KOH = volume (in L) x concentration (in mol/L)
= 75.0 mL x (1 L / 1000 mL) x 0.15 mol/L
= 0.01125 mol
Number of moles of HCN = volume (in L) x concentration (in mol/L)
= 35.0 mL x (1 L / 1000 mL) x 0.20 mol/L
= 0.0070 mol
Since HCN has fewer moles (0.0070 mol) compared to KOH (0.01125 mol), HCN is the limiting reagent.
Step 3: Calculate the reaction quotient (Q):
The reaction quotient is calculated in the same way as the equilibrium constant, but we use the initial concentrations instead.
Q = [C]^c × [D]^d / [A]^a × [B]^b
Since KOH is in excess, its concentration does not change. Therefore, [KOH] = 0.15 M
For HCN, the concentration after the reaction will be:
[HCN] = [initial HCN] - [consumed HCN]
= 0.20 M - 0.0070 M
= 0.193 M
Since KCN is formed in the reaction, its concentration after the reaction will be equal to the reacted HCN concentration:
[KCN] = [consumed HCN]
= 0.0070 M
For water (H2O), we assume its concentration remains constant since it is a pure liquid.
Plugging these values into the reaction quotient equation gives us:
Q = [KCN] / ([KOH] × [HCN])
= 0.0070 M / (0.15 M × 0.193 M)
= 0.3202
Step 4: Determine the direction of the reaction:
Since Q = 0.3202 and Ka = 4.9 × 10^–10, we can compare the reaction quotient to the equilibrium constant:
If Q > Ka, the reaction proceeds from right to left (towards the reactants).
If Q < Ka, the reaction proceeds from left to right (towards the products).
If Q = Ka, the reaction is at equilibrium.
In this case, Q > Ka, which means the reaction proceeds towards the reactants (from right to left).
Step 5: Calculate the concentration of H3O+ ion:
Since the reaction shifts to the left, some HCN will react with water to form hydronium ions (H3O+). The concentration of H3O+ can be determined using the initial concentration of HCN and the stoichiometric ratio between HCN and H3O+ in the balanced equation:
[HCN] = [H3O+]
Therefore, [H3O+] = 0.193 M
Step 6: Calculate the pH:
pH is a measure of the acidity of a solution and is defined as the negative logarithm (base 10) of the hydronium ion concentration:
pH = - log [H3O+]
pH = - log (0.193)
= 0.716
Therefore, the pH of the resulting solution is approximately 0.716.