The set up from part A looks like this.
1g/74.08 = moles B = 0.0135
1/110.54 = moles BH^+ = 0.009046
YOu had 100 mL soln for part A; therefore, we have 13.5 mmoles B and 9.046 mmols BH^+
From part c we add 5 mL x 0.1M = 0.5 mmoles.
............B + H^+ ==> BH^+
initial...13.5..0.......9.046
add............0.5.............
change....-0.5..-0.5.....+0.5
equil.....13.0...0.......9.546
pH = 8.2 + log(13.0/9.546) = 8.33
The others are done the same way.
Glycine amide hydrochloride (BH+) FM 110.54, pKa 8.20
Glycine amide (B) FM 74.08
b.) How many grams of glycine amide should be added to 1g of glycine amide hydrochloride to give 100ml of solution with pH 8?
I tried using the henderson-hasselbalch equation to give:
8 = 8.2 + log((B/BH+)) though now I'm not sure exactly what I'm trying to solve for.
C.) What would be the pH if the solution in (a) is mixed with 5mL of 0.100M HCl?
Part (a) refers to a solution prepared by dissolving 1g glycine amide hydrochloride plus 1g of glycine amide in 0.100L.
Answer is 8.33
D.) What would be the pH if the solution in (c) is mixed with 10mL of .100M NaOH?
Answer is 8.41
As always, I just need help on finding out how to get to the answers.
Thank you.
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