For 270.0 mL of a buffer solution that is 0.245 M in HCHO2 and 0.305 M in KCHO2, calculate the initial pH and the final pH after adding 0.005 mol of NaOH.
Concentration of HCHO2 = 0.245M
Concentration of KCHO2 = 0.305M
Volume of the buffer solution = 270.0 mL = 0.270L
number of moles = Molarity x volume of the solution in L
number of moles of HCHO2 = 0.245M x 0.270L = 0.06615 moles
number of moles of KCHO2 = 0.305M x 0.270L = 0.08235 moles
Ka of HCHO2 = 1.8 x10^-4
Ka = 1.8 x10^-4
-log(Ka) = -log( 1.8 x10^-4)
PKa = 3.74
At initial
PH = PKa + log[salt]/[acid]
PH = Pka + log[KCHO2]/[HCHO2]
PH = 3.74 + log( 0.08235/0.06615)
PH = 3.83
after addition of NaOH
number of moles of NaOH = 0.005 moles
after addition of NaOH
number of moles of HCHO2 = 0.06615 - 0.005 = 0.06115 moles
number of moles of KCHO2 = 0.08235 + 0.005 = 0.08735 moles
PH = Pka + log[salt]/[acid]
PH = 3.74 + log( 0.08735/0.06115)
PH = 3.89
For 270.0 mL of a buffer solution that is 0.245 M in HCHO2 and 0.305 M...
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