A titration curve was obtained by titrating 25ml boric acidsolution with a 1M Na

Question

A titration curve was obtained by titrating 25ml boric acidsolution with a 1M NaOH solution. Use the information given on thecurve to answer the following:
What was the concentration of the original boric acid solution?
Given:
Curve with inital acid pH equal to 4.17 0mL NaOH added, pH athalfway point is 9.137 with 10ml NaOH added, and pH at equivelencepoint is 11.40 with 20ml NaOH added. And calculated Ka for boricacid to be 7.29*10-10

Explanation / Answer

Chemical reaction:                                  H3BO3 (aq)   +   NaOH (aq) -------------->  NaBO2 (aq) + 2 H2O (l)             One mole of boric acid reacts with one mole of base.      Initial pH = 4.17                                  B(OH)3 (aq) + H2O (l)    B(OH)4-(aq)   + H+ (aq)                             [H+]   = 6.8 * 10-5 M   =   [B(OH)4-]                             [B(OH)3 ] = 0.8 M - 6.8 *10-5 M                                               ˜ 0.8 M                                      Ka   = 7.29*10-10                          7.29*10-10 = 6.8 * 10-5 *6.8 * 10-5   /x                                          x = 6.8 * 10-5 * 6.8 *10-5 / 7.29*10-10                                                              = 6.34                           This is the equilibrium concentration of B(OH)3 .Initially it will be 6.34 M + 6.8 * 10-5 M.                                                                                                                          ˜ 6.340068 M

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