Suppose that 25. 0 ml of 0. 10 m ch3cooh(aq) ( ka =1. 8 x 10-5) is titrated with 0. 10 m naoh(aq). (a)what is the initial ph of
1 answer:
The initial ph of the 0. 10M ch3cooh (aq) solution is -0.64.
The salt sodium acetate is present at the equivalency point. It is a salt of a strong base and a weak acid. The expression "its pH" provides the information.
pH = 7 + 0.5p + 0.5logC
where C is the salt concentration. Monobasic acid is Acetic acid here.
25 ml of 0.10 M NaOH solution are needed for the titration of 25 ml of 0.10 M acetic acid.
Total volume = 25 + 25 = 50ml
The salt concentration = C = 0.10× = 0.05
The pH for weak acid is given by,
pH = (p
+ logC)
pH = (1.8× + log0.05)
pH = [1.8× + (-1.30)]
pH = (1.8× -1.30)
pH = -0.64
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Answer : The molar mass of the solute will be 87.90 g/mol.
Explanation : We know the formula for elevation in boiling point, which is
Δt = i
m
given that, Δt = 0.357, = 5.02 and mass of 
= 40,
on substituting the value we get,
0.357 = (1) X (5.02) X (x/ 0.044), on solving we get x = 2.844 X
.
Now, 0.250/ 2.844 X = 87.90 g/mol. which is the weight of unknown component.
Explanation : We know the formula for elevation in boiling point, which is
Δt = i
given that, Δt = 0.357,
on substituting the value we get,
0.357 = (1) X (5.02) X (x/ 0.044), on solving we get x = 2.844 X
Now, 0.250/ 2.844 X