calculate the pH of the buffer that results from mixing 51.5 mL of a 0.398 M solution of HCHO2 and 11.5 mL of a 0.605 M solution of NaCHO2. the Ka value for HCHO2 is 1.8 x 10^-4
Buffer solution is a solution which resists the change in pH.
Buffer solutions are of two types.
1. Acidic buffer: It is a mixture of weak acid and salt formed by this weak acid with a strong base.
Ex: Acetic acid and sodium acetate (salt of acetic acid with NaOH )
2. Basic buffer: It is a mixture of weak base and salt formed by this weak base with a strong acid.
Ex : Ammonium hydroxide and Ammonium chloride
The given buffer is an acidic buffer ( formic acid and sodium formate)
The pH of buffer is given by Henderson's equation.
Or
Molarity is the number of moles of solute present in one litre of the solution.
M = Number of moles of solute * 1000/Volume of the solution
Number of moles of solute = M * Volume of solution in mL /1000
Number of moles of formic acid in 51.5 mL of 0.398 M formic acid solution is
Number of moles of formic acid =
0.398 * 51.5/1000
= 0.0205
Number of moles of sodium formate in 11.5 mL of 0.605 M sodium formate is
Number of moles of sodium formate =
0.605 * 11.5/1000
= 0.00696
Given that Ka of formic acid = 1.8×10-4
pKa = -logKa = -log(1.8×10^(-4))
= 4 -log1.8
= 4 - 0.255 = 3.745
pH of the given buffer is
= 3.745 + log(0.339)
= 3.745 - 0.469 = 3.276
calculate the pH of the buffer that results from mixing 51.5 mL of a 0.398 M...
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