I need help with two different problems. I think I may have solved them, but need some confirmation that I did them correctly:
Problem 1:Calculate the molarity of NaOH solution which requires 44.62 mL to titrate 1.730 grams of thepure oxalic acid, H
2C
2O
4.2H
2O to the phenolphthalein
end point.My attempt:On this problem I converted oxalic acid to moles= 3.07x10
-1 M H
2C
2O
4.2H
2O
In calculating the end point it is my understanding that you need [I-]/[HI]=1/10 and setting this to [NaOH]/(3.07x10
-1) to solve for [NaOH]. I did this and got 3.7x10
-2 M NaOH as the final answer. Is this correct?
Problem 2:A weak diprotic acid has a cesium salt, CsHA which is very slighlty soluble. You titrate 100.0 mL of the saturated solution with 0.0499 M NaOH and reach an endpoint at 17.30 mL. One of your data points is at 8.30 mL and the pH recorded for it is 4.88. What is the solubility product, K
sp, and the K
a2 for this compound CsHA?
My attempt:Again, I tried solving for it and used M
aV
a=M
bV
bV
a= 100.0 mL
M
a=0.0499
V
b=8.30 mL
and my answer yielded 4.14x10
-3 M CsHA
I then set up the equation as CsHA + H
2O
H
3O + CsA
-[H]= 10
-4.88 = 1.32x10
-5Ksp = [H
30
+][CsA] = (1.32x10
-5)2=1.74x10
-10For Ka I used H&H equation where 4.88=pK
a + log(8.30/(17.3-8.3))
pK
a=4.92 and K
a=1.2x10
-5Does this look like I set it up correct?
Thanks