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Question 32 (1 point) What is the pH of a 0.056 M aqueous acetic acid solution? kg = 1.8 x 10-5. Your Answer: Answer

Answer 1

Question 32:

Given that,

concentration of acetic acid is C = 0.056 M,

and the Ka = 1.8×10^-5.

We have to calculate the pH.

So, pKa = -logKa = -log(1.8×10^-5) = 4.74.

Hence, the required pH is

pH = (1/2)[pKa - logC] = 2.99 = 3.0.

Thus, the required pH is pH = 3.0.

Question 31.

Given that, k_{​​​​​​f} = 1.86 ^oC/m.

W2 = 72 g, W1 = 629 g and M2 = 46.08 g/mol.

We have to calculate the depression of freezing point.

So, molality is m = [1000×W2/M2×W1] = 2.484 m.

Hence, the depression of freezing point is

$\Delta$ T_f = k_f.m = 4.62 ^oC.

Therefore the freezing point is lower by 4.62 ^o​​​​​​C.

Question 30.

Both the CH2Cl2 and CF2Cl2 are polar molecules.

Hence the correct option is the third one i.e. both are polar.

Question 29.

The formal charge on the nitrogen atom in the given molecule is +1.

Thus the correct option is the third one i.e. +1.