Answer:
0.1795 M
Explanation:
From the given information:
The equation is:
[tex]\mathbf{HF_{(aq)}+ KOH_{(aq)} \to KF _{(aq)} + H_2O}[/tex]
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)
[tex]\text{Number of moles = Volume} \times Molarity[/tex]
Thus. since moles of HF = moles of KOH
Then; 0.020 × 0.393 M = 0.02379 × (x) M
(x) M = [tex]\dfrac{0.02 \times 0.393 \ M}{0.02379}[/tex]
(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 = [tex]\dfrac{moles}{volume}[/tex]
Molarity = [tex]\dfrac{0.00786 \ mol}{0.04379 \ L}[/tex]
Molarity = 0.17949 mol/L
= 0.1795 M
After the initial neutralization, the molarity at the equivalence point will be:
"0.17949 mol/L"
According to the question,
→ HF (aq) + KOH (aq) [tex]\rightarrow[/tex] KF (aq) + H₂O
Since, moles of HF = moles of KOH
We know that,
Number of moles = Volume × Molarity
→ 0.020 × 0.393 = 0.02379 × M
By applying cross-multiplication,
The molarity of KOH be:
M = [tex]\frac{0.02\times 0.393}{0.02379}[/tex]
= [tex]\frac{0.00786}{0.02379}[/tex]
= 0.3304 M
By using balanced neutralization reaction,
Volume = 0.020 + 0.02379
= 0.04379 L
hence,
The molarity will be:
= [tex]\frac{Moles}{Volume}[/tex]
By substituting the values,
= [tex]\frac{0.00786}{0.04379}[/tex]
= 0.17949 mol/L
Thus the response above is correct.
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