Acid and base behavior can be observed in solvents other than water. One commonl

Question

Acid and base behavior can be observed in solvents other than water. One commonly used solvent is dimethyl sulfoxide (DMSO), which can be treated as a monoprotic acid "HSol." Just as water can behave either as an acid or a base, so HSol can behave either as a Brønsted-Lowry acid or base.

A)The equilibrium constant for self-dissociation of HSol (call it { m K}_{{ m H}}Sol) is 1×1035. Write the chemical equation for the self-dissociation reaction and the corresponding equilibrium equation. (Hint: The equilibrium equation is analogous to the equilibrium equation for Kw in the case of water.)

B) The weak acid HCN has an acid dissociation constant Ka=1.3×1013 in the solvent HSol. If 0.010 mol of NaCN is dissolved in 1.00 L of HSol, what is the equilibrium concentration of { m H}_{2}Sol+?

Explanation / Answer

A)

Self Dissociation reaction for Water (H2O) is given as,

H2O + H2O <---------> H3O+ + HO-.

Analogously we write Self Dissociation reaction for DMSO (HSol) s,

HSol + HSol <-------> H2Sol+ + Sol- (aq).

K = [H2Sol+][Sol-]/[HSol]2 = 1 x 10-35.

K x [HSol]2 = [H2Sol+][Sol-] = 1 x 10-35.

[HSol] is almost unchanged (i.e. constant) as HSol is in excess amount. Hence we write

KDMSO = [H2Sol+][Sol-] = 1 x 10-35. (where, KDMSO =Kx[HSol]2).

KDMSO = [H2Sol+][Sol-] = 1 x 10-35.

Is the required expression.

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B) NaCN is a strong salt and ionizes completely in DMSO phase as,

NaCN -----------> Na+ + CN-

So, [CN-] = [NaCN] = 0.01 moles/L = 0.01 M

C = 0.01 M.

pKa = -log(1.3x10-13) = 12.89

KDMSO = 1 x 10-35.

So, p(KDMSO) = -log(1 x 10-35) = 35

p(H2Sol+) = ?

Formula for ,

p(H2Sol+) = (1/2) x [pKDMSO + pKa) + log(c)]

p(H2Sol+) = (1/2) x [35 + 12.89 + log(0.01)]

p(H2Sol+) = (1/2) x [35 + 12.89 + (-2)]

p(H2Sol+) = (1/2) x [45.89]

p(H2Sol+) = 22.95

By definition,

p(H2Sol+) = -log[H2Sol+])

22.95 = -log[H2Sol+])

[H2Sol+] = 10-22.95.

[H2Sol+] = 1.12 x 10-23 M

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