What is the pH of a buffer prepared by mixing 182 mL of 0.442 M HCl an

QUESTION

What is the pH of a buffer prepared by mixing 182 mL of 0.442 M HCl and 0.530 L of 0.400 M sodium acetate. (Acetic Acid: pKa = 4.74)
So theres 2 things going on here. HCl is dissociating leaving some H in the solution, and the sodium acetate is dissociating, then equilibrating with the H . So first off, figure out how much H there is. Since were dealing with HCl, its pretty easy: 182 mL HCl * .442 M * 1 mmol H /1 mmol HCl = 80.4 mmol H Then figure out the mmol of acetate: 530 mL sodium acetate * .400 M * 1 mmol Ac “ / 1 mmol sodium acetate = 212 mmol Ac “ These are basically going to react to form HAc then equilibrate. LR is obviously H , so 132 mmol of Ac “ is left over (strictly following sig fig rules). Now we can¦

make the ice chart for the equilibrium reaction: (remember to find concentration, not just put in mmol) HAc <---> H Ac “ 0.113 0 0.185 -x x x 0.113-x x 0.185 x Now we need K a , which is just 10^-pk a , so 10^-4.74 = 1.82E-5 Now set up the K a expression and solve for x: 1.82E-5 = ((x)(0.185 x))/(0.113-x), x=1.11E-5 And since thats the concentration of H , pH is just -log(1.11E-5) = 4.95

ANSWER:

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