You have a flask in your laboratory on which the label is smudged. From what you

Question

You have a flask in your laboratory on which the label is smudged. From what you are able to read, it is a solution of a monoprotic acid (we’ll call it HX) at a concentration of 2E-3 M in a 0.05 M NaCl solution. You measure the pH of the solution and get a reading of 3.09. Using this information, do the following:

(a) Determine the pKa of the acid and the identity of the X- using the information in Table 8.2. Recall that a pH meter measures activity, not concentration of protons.

272 Water Chemistry Table 8.2 Acidity constants (pKa) for some common acids * Values selected to agree with MINEQL + database where possible

Explanation / Answer

a) Ok, as you have an acid pH, we'll assume this is a strong acid. NaCl doesn't contribute to pH because Na+ and Cl- just dissociate in water, they don't react.

Ka = [X-][H3O+] / [HX] ; pH = 3.09 ---> [H3O+] = 8.13*10-4 = x

Ka = x2 / (2E-3 - x)

Ka= (8.13*10-4 )2 / (2E-3 - 8.13*10-4 ) = 5.57 E-4

pKa= -logKa = 3.254 ------> HF is closer to this.

b) I think you ment an HX solution:

Ka = x2 / ([HX] -x)

8.13*10-4 = x2 / (4E-3 -x)

x= 1.8 E-3 = [H3O+]

pH = -log (1.8E-3) = 2.74

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