HARDMCQ SINGLENEET ChemistryPhysical ChemistryIonic equilibriumAchiever Section
A student dissolved 0.100 mol of an unknown monoprotic acid, HA, in sufficient water to make 1.00 L of solution. She measured the pH of the solution as pH = 2.60. Of the following, the acid HA is
A.acetic acid Ka = 1.8 × 10–5mol L–1
B.benzoic acid Ka = 6.6 × 10–5mol L–1✓ Correct
C.hypochlorous acid Ka = 3.1 × 10–8mol L–1
D.hydrofluoric acid Ka = 3.5 × 10–4mol L–1
Explanation
The concentration of the monoprotic acid ([HA]) will be the number of moles divided by volume and here it will be 0.1 HA 0.1M 1 . Now, pH= log H and here pH=2.6. So, 2.6 3 2.6 log H H 10 H 2.5 10 >> >> x We know that for weak acid, H c , where c = concentration of the acid and = degree of dissociation. So, H c H c >> Also, the value of the acid dissociation constant is 2 a K c . So, the substituting the values we get, 2 a 2 3 a 6 a 5 5 a H K c c 2.5 10 K c c 6.25 10 K 0.1 K 6.25 10 6.6 10
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