The molality of the solution is obtained as 0.63 m.
<h3>What is the freezing point?</h3>
The freezing point is the temperature at which the liquid is converted into solid.
We know that;
ΔT = 3.5° C
K = 1.86° C/m
i = 3
m = ?
Thus;
ΔT = K m i
m = ΔT/K i
m = 3.5° C/ 1.86° C/m * 3
m = 0.63 m
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Answer:
The pH of the buffer solution = 8.05
Explanation:
Using the Henderson - Hasselbalch equation;
pH = pKa₂ + log ( [HPO₄²-]/[H₂PO4⁻]
where pKa₂ = -log (Ka₂) = -log ( 6.1 * 10⁻⁸) = 7.21
Concentration of OH⁻ added = 0.069 M (i.e. 0.069 mol/L)
[H₂PO4⁻] after addition of OH⁻ = 0.165 - 0.069 = 0.096 M
[HPO₄²-] after addition of OH⁻ = 0.594 + 0.069 = 0.663 M
Therefore,
pH = 7.21 + log (0.663 / 0.096)
pH = 7.21 + 0.84
pH = 8.05
Explanation:
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