A 15 mL solution of 0.03 M HCI is titrated with 0.03 M NaOH to give the followin

Question

A 15 mL solution of 0.03 M HCI is titrated with 0.03 M NaOH to give the following titration curve. Determine the key features of this curve. (The titration curve is approximate and may not show all aspects adequately.) The titration curve has the following features. Before any base is added, the pH of the starting solution of 0.03 M HCI is When small amounts of base are added, up to about 13 mL, the pH gradually but remains in the region. In this area, the solution contains an excess of H_3O^+ ions. At and around the equivalence point, the pH rises dramatically, going from the acidic region to the basic region. At the equivalence point, when [H_3O^+] = [OH^-], the pH is As more base is added beyond the equivalence point, the pH gradually within the region. In this area, the solution contains an excess of OH^- ions.

Explanation / Answer

1. Before adding base, pH = -log[H+]

pH = -log[0.03] = 1.52

2. A/c to MaVa = MbVb

MaVa = 15x 0.03 =0.45; MbVb = 13 x0.03 = 0.39

MaVa >MbVb, so the soluton is acidic

hence [H+]= MaVa - MbVb/Va+Vb

. = 0.45 - 0.39/28 = 0.0021

pH = - log(0.0021) = 2.67 and pH is inceased and solution remains acidic.

3. At the equivalence point Vb= 15 mL, then MaVa = MbVb, neutral solution. pH = 7

4. pH increases gradually as the base added and solution is in basic regeion.

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