Can someone help me with the following questions if they have a bit of spare time 
I am really struggling with these ones!
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Which of the following mixtures would result in a buffered solution when 1.0 L of each of the two solutions are mixed?
a) 0.2 M HNO3 and 0.4 M NaNO3
b) 0.2 M HNO3 and 0.4 M HF
c) 0.2 M HNO3 and 0.4 M NaF
d) 0.2 M HNO3 and 0.4 M NaOH
Answer is C, but I don't understand why
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Calculate the number of moles of HCl(g) that must be added to 1.0 L of 1.0 M NaC2H3O2 to produce a solution buffered at
a) pH = 4.2
b) pH= 5.0
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A solution contains 1.0 x 10^-6 M HOCl and an unknown concentration of KOCl. If the pH of the solution is 7.20, calculate the KOCl concentration.
(Hint; The contribution of water to the [H+] cannot be ignored.)
I keep getting the answer 5.5 x 10^-7M. But the answer is 7.0 x 10^-7M, and I am not sure why this is the case.
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Consider a weak acid, HA, with a Ka value of 1.6 multiplied by 10-7. Calculate the pH of a solution that is 5.0 multiplied by 10-7 M HA and 5.0 multiplied by 10-7 NaA.
I am trying to use the Henderson Hasselbach Equation here, but it keeps giving me the wrong answer. I keep getting Ph=6.8.
First question is interesting.
a is wrong because that's a strong acid and its conjugate base, essentially a strong acid solution
b is wrong because it's just two acids
d is wrong because it's a strong acid and a strong base and after neutralisation, you have more base left
Why is C correct? After neutralising 1 lot of H+ with 2 lots of F-, you have 1 lot of F- and 1 lot of HF which is exactly what a buffer is: a weak acid and its conjugate weak base in equal amounts.
2. Using the HH equation, for a solution of pH 4.2, given Ka of acetic acid 1.8*10^-5:
After the reaction with base you have 10^-4.2 M H+ due to the given pH
Before the reaction you have 10^-4.2 + x M H+ if x is the concentration of acetic acid present at equilibrium, as well as 1.0 M base. So afterwards you have 1-x M base
At equilibrium, 10^-4.2 * x/(1-x) = 1.8*10^-5
Solve for x. This will be the concentration of acetic acid at equilibrium. Given that you don't initially have any acetic acid to begin with (by assumption), this will have come from the addition of acid. Hence x in M is the concentration of HCl added => work out number of moles
Do the same thing except with pH 5.0 instead of 4.2 for the second part
3. H2O <=> H+ + OH-
HOCl <=> H+ + OCl-
Let the initial concentration of KOCl be x. Assume initially you have 10^-7 M acid and base. Let the amount of H+ formed be y. [OCl-] at equilibrium is x + y
Then, (10^-7 + y)*[OH-] = 10^-14
(10^-7 + y)(x + y)/(1 - x) = 3.5 * 10^-8
(10^-7+y)(x) + y(10^-7+y) = 3.5*10^-8 - 3.5x*10^-8
x = (3.5*10^-8-y(10^-7+y))/(10^-7+y+3.5*10^-8)
Given that [H+] = 10^-7.2 = 10^0.8 * 10^-8, y = 10^-8*(10^0.8 - 10), plugging this into the second equation gives you your answer (I don't have a decent calculator to solve this)
4. For this one, take into account the dissociation of water as well. Start with [H+] = [OH-] = 10^-7 M and set up your ICE equations based upon that. HH doesn't work well here
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