Question

Sonni added 23.79 mL of KOH to 20.00 mL of 0.393 M HF until the equivalence point was reached. What is the concentration of conjugate base at the equivalence point after the initial neutralization?

Answer 1

Answer:

0.1795 M

Explanation:

From the given information:

The equation is:

$\mathbf{HF_{(aq)}+ KOH_{(aq)} \to KF _{(aq)} + H_2O}$

From the above equation, the reaction will go to completion due to the strong base:

At the equivalence point, moles of acid (HF) will be equal to moles of base (KOH)

$\text{Number of moles = Volume} \times Molarity$

Thus. since moles of HF = moles of KOH

Then; 0.020 × 0.393 M = 0.02379 × (x) M

(x) M = $\dfrac{0.02 \times 0.393 \ M}{0.02379}$

(x) = 0.3304 M

Thus, the molarity of KOH = 0.3304 M

Using the balanced neutralization reaction;

moles of HF = moles of KOH = moles of conjugate base = 0.00786 mol

∴

Volume = 0.020 L + 0.02379 L

Volume = 0.04379 L

Volume of the solution =  0.04379 L

Therefore; Molarity = $\dfrac{moles}{volume}$

Molarity = $\dfrac{0.00786 \ mol}{0.04379 \ L}$

Molarity = 0.17949 mol/L

= 0.1795 M