Answer:
The pH of the solution is 3.23
Explanation:
The reaction between HF and KOH is:
KOH + HF → KF + H₂O
The number of moles of HF in 100.0 mL is:
1000.0 mL -------------- 0.20 mol
100.0 mL --------------- x= 0.02 mol
The number of moles of KOH in 75.0 mL is:
1000.0 mL -------------- 0.10 mol
75.0 mL ----------------- x= 7.5×10⁻³ mol
We work with moles because when two solutions are mixed, the solution volume increases. As the volume increases, molarity will change but the number of moles will remain the same.
The changes in the number of moles are:
KOH + HF → KF + H₂O
Initial (mol) 7.5×10⁻³ 2.0×10⁻² 0
Change (mol) -7.5×10⁻³ -7.5×10⁻³ 7.5×10⁻³
Final (mol) 0 1.25×10⁻² 7.5×10⁻³
At this stage we have a buffer system made up of HF and F⁻ (from the salt, KF). To calculate the pH of the solution, we write:
Ka= [H⁺][F⁻] / [HF]
[H⁺]= [HF]Ka / [F⁻] = 5.83×10⁻⁴
Therefore, pH= 3.23
The pH of the resulting solution is 1.15.
The equation of the reaction is;
HF(aq) + KOH(aq) ------> KF(aq) + H2O(l)
Number of moles of HF = 100.0/1000 L × 0.20 M = 0.02 moles
Number of moles of KOH = 75.0/1000 L × 0.10 M = 0.0075 moles
Clearly, the acid is in excess hence we have to obtain the concentration of the excess acid.
Number of moles of excess acid = 0.02 moles - 0.0075 moles = 0.0125 moles
Total volume of solution = 100.0 mL + 75.0 mL = 175 mL or 0.175 L
Concentration of excess acid = 0.0125 moles/0.175 L = 0.07 M
pH = -log[H^+]
pH = -log[ 0.07 M]
pH = 1.15
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