5.Using the equation you have just derived in four above, calculate the pH of a solution composed of 155 mL of 0.1 M acetic acid and 45 g of sodium acetate. The pKa of acetic acid is 4.76.
First find the moles of sodium acetate:
45g x 1mol/60g = .75mol
Then divide that by the total amount of solution to find the concentration:
.75mol/.155L = 4.84 M
Then plug the concentrations and the pKa into the equation:
pH = 4.76 + log ([4.84]/[0.1])
pH= 6.44
6.What would the pH of this above solution, from five, be after 0.01 mole of HCl was added?
I have no idea how to do this, I don't even know if I got 5 right. I tried.
2007-03-04 10:11:22 · 1 answers · asked by Anonymous in Science & Mathematics ➔ Chemistry
So I got for number 6. pH= 6.8..... Is this right?
2007-03-04 10:59:50 · update #1
You've got the Mr of sodium ethanoate wrong. It's 82. Otherwise your method is correct.
Now add 0.01 moles to the ethanoate moles, and take 0.01 off the acid moles. Now work out their new molarities, and "plug them into the equation". You should find that the pH has hardly changed.
2007-03-04 10:16:40 · answer #1 · answered by Gervald F 7 · 1⤊ 0⤋