1. What is the pH of a buffer solution that is 45% acetic acid and 55% sodium acetate?
2. When an interfering substance M is present, a fluoride ion selective electrode obeys the following modified response equation:
E(V) = K - β(0.0592)log{[F-]+Ks[M-]}
A potential of -277.0 mV is measured for a fluoride ion selective electrode in a solution of 1.00×10-5 M fluoride at pH 5.00. Calculate the potential that would be measured if the solution pH were 9.90, assuming the selectivity coefficient (Ks) for OH- is 0.1 and that β is 1.00.
Even if you only know how to do one, anything helps. THANKS!
2. When an interfering substance M is present, a fluoride ion selective electrode obeys the following modified response equation:
E(V) = K - β(0.0592)log{[F-]+Ks[M-]}
A potential of -277.0 mV is measured for a fluoride ion selective electrode in a solution of 1.00×10-5 M fluoride at pH 5.00. Calculate the potential that would be measured if the solution pH were 9.90, assuming the selectivity coefficient (Ks) for OH- is 0.1 and that β is 1.00.
Even if you only know how to do one, anything helps. THANKS!
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1. pH = pKa + log [salt]/[acid] Here [salt] = 55/82 and [acid] = 45/60 and pka= 4.78
Mol. wt. sod acetate = 82 and that of acetic acid =60
2. Calculate 'K' from given data
then beta = 1.0, [F-] = 1 x10^-5 , Ks = 0.1 and calculate [H+] from pH 9.9 and obtain [OH-] = 10^14/ [H+]
Find E
Mol. wt. sod acetate = 82 and that of acetic acid =60
2. Calculate 'K' from given data
then beta = 1.0, [F-] = 1 x10^-5 , Ks = 0.1 and calculate [H+] from pH 9.9 and obtain [OH-] = 10^14/ [H+]
Find E