To calculate the pH of the solution resulting from mixing NaOH and HF, we first need to determine the moles of each substance present after mixing.
Given:
- Volume of NaOH solution (V1) = 122
mL
- Concentration of NaOH solution (C1)
= 0.150 M
- Volume of HF acid (V2) = 75.5 mL
- Concentration of HF acid (C2) =
0.120 M
Step 1: Calculate the moles of NaOH and HF:
Moles of NaOH (n1) = C1 * V1 / 1000
(Convert mL to L by dividing by 1000)
Moles of NaOH (n1) = 0.150 mol/L *
122 mL / 1000 = 0.0183 moles
Moles of HF (n2) = C2 * V2 / 1000
Moles of HF (n2) = 0.120 mol/L * 75.5
mL / 1000 = 0.00906 moles
Step 2: Determine the limiting reactant:
To find the limiting reactant, compare the moles of NaOH and HF. The reactant that is present in lower moles is the limiting reactant.
In this case, HF is the limiting reactant because it has fewer moles compared to NaOH.
Step 3: Calculate the excess amount of NaOH left after the reaction:
Since NaOH is in excess, the moles of NaOH remaining after the reaction will be equal to the initial moles of NaOH.
Step 4: Calculate the total volume of the solution after mixing:
Total volume (V_total) = V1 + V2 =
122 mL + 75.5 mL = 197.5 mL
Step 5: Calculate the concentration of the resulting solution:
Total moles = n1 + n2 (since HF is
limiting)
Total moles = 0.0183 moles +
0.00906 moles = 0.02736 moles
Concentration of the resulting solution:
C_ final = Total moles / V_total
C_final = 0.02736 moles / (197.5 mL /
1000) = 0.1385 M
Step 6: Calculate the pOH of the solution:
pOH = -log[OH-]
pOH = -log(0.1385)
рОН = 0.859
Step 7: Calculate the pH of the solution:
pH + pOH = 14
pH = 14 - pOH
pH = 14 - 0.859
pH = 13.141
Therefore, the ph of the resulting solution is approximately 13.141.
