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2. Calculate the pH of a buffer solution formed by dissolving 0.350 mole HCH3COO and 0.550 mole of NaCH3COO in 0.950L solution. Ka HCH3COO = 1.8 x 10-6. The buffer in the solution is acid buffer made up of HCH3COO Weak acid [A-] CH3COO- Salt of the weak acid [HA]​

Sagot :

Explanation:

given:

Ka: 1.8x10^-6

NaCH3COO = 0.550 mole

HCH3COO = 0.350 mole

Solution = 0.950L

first, get the molar concentration of NaCH3COO & HCH3COO in the solution.

NaCH3COO: 0.550mole/0.950 L = 0.578 M

HCH3COO: 0.350mole/0.950 L= 0.368 M

second, compute for pKa

pKa= -logKa

pKa= -log( 1.8x10^-6

pKa= 5.744

third, subsitute to hendersson-hasslebach equation

pH = pKa +log [A-] / [HA]

pH= 5.744 +log( 0.578 / 0.368 )

pH = 5.94