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Consider the titration of 50.0 mL of 0.5 M weak acid analyte HF (Ka = 6.6...

Consider the titration of 50.0 mL of 0.5 M weak acid analyte HF (Ka = 6.6 x 10-4) with 1.0 M titrant NaOH. A. (5 pts) What is the volume of base added at the stoichiometric (equivalence) point?

Solutions

Expert Solution

[H+] ion concentration of weak acid = (Ka*C)^1/2

                                                    = (6.6*10^-4 *0.5)^1/2

                                                     = 1.82*10^-2

Now we have to use M1V1 = M2V2

     M1 = 1.82*10^-2 , V1 = 50 ml , M2 = 1.0 M , V2 = ?

    1.82*10^-2*50.0 = 1.0*V2

      V2 = 0.91 ml

volume of base added is 0.91 ml

queen_honey_blossom answered 2 years ago
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