Calculate the amount of acetic acid and sodium acetate needed to prepare 100.0 mL of acetic buffer solution (0.100M) with pH = 5.10 by using a solution of acetic acid (25% w/w, density = 1.05 g/cm3) and pure sodium acetate. (pKa = 4.76; M.W. = 60.05 g/mol)
Start by using the Henderson-Hasselbalch equation, then find out the moles of base and acid required. Use the solutions you were given to achieve these values.
Is the equation Hc2H3O2(aq) + H2O (l) <---> C2H3O2- (aq) + H3O+ (aq)?
For the dissociation of the acid, yes
It's also the one that would work here since you are using the pKa
So then am I solving for A-?
no, you're solving for the ratio of conjugate base/acid
so a-/ha?
that is what you are solving for first
Then you have to find the moles of each (acid and conjugate base) based on the percentage
I'm a little rusty on basic math, but to cancel log on one side, you put both sides to the power of ten, right? Like 10^(x)?
yup thats right
And as you said, step 2 is (a)/(HA)*0.25= Acetic acid?
not quite. \(\dfrac{[A^-]}{[HA]}=10^{(pH-pKa)}\) Then find how much (i.e. percentage) of 0.100M each needs to have
These examples make it clear https://www.researchgate.net/file.PostFileLoader.html?id=54f9c88fd11b8b402a8b4586&assetKey=AS%3A273725033779218%401442272541972
Btw, if you click that link it will download a pdf
"*Since [A-] / [HA] = 1.584893192, we can say that [A- ] / [HA] = 1.584893192/ 1. In this case [A-] = 1.584893192; [HA] = 1." I don't understand that asumption.
It's because of the fraction. you have \(\dfrac{[A^-]}{[HA]}=2.187761\) which is\(\dfrac{[A^-]}{[HA]}=\dfrac{2.187761}{1}\) (any number can be written with a 1 in the denominator) so you have \([A^-]=2.187761\) and \( [HA]=1\) The you find the percent: percent of \([A^-]=\dfrac{[A]}{[A]+[HA]}=\dfrac{2.187761}{2.187761+1}\)
Ok, that makes sense.
Okay cool
And I multiply %{a-} with 100.0 mL acetic buffer solution?
If you follow that pdf, youll see that you first multiply by the molarity of the desired solution, then by the volume (in L) of the solution to find the moles needed.
When I get to the point of calculating the liters, by what value of moles are A- and HA multipled/divided by?
There's only one provided molarity, which is for the buffer solution (0.0100 M).
So if you're following that pdf, you would multiply the percentage of, both, A^- and HA by 0.1 M, then by 0.1 L. This gives you the moles of each that the buffer needs. Now you would use the rest of the information to find how much of each solution/solid you would need. Find the molarity of the acetic acid solution (25% w/w, density =1.05 g/cm3) and use it to find the volume of solution that contains that many moles of Acetic acid. Similarly, for Na acetate, find the moles you need. (this one is a solid, so you only need molar mass and your value will be in grams)
I have to go, I hope you can finish this by yourself, or maybe someone else can continue helping you. good luck!
To anyone who sees this, I'm stuck on, " Find the molarity of the acetic acid solution (25% w/w, density =1.05 g/cm3) and use it to find the volume of solution that contains that many moles of Acetic acid.". How do I cancel out cm^3 and get L from this?
Anyone know how significant digits would factor here?
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