posted by Andrew .
Calculate the pH at the halfway point and at the equivalence point for each of the following titrations.
(a) 100.0 mL of 0.14 M HC7H5O2 (Ka= 6.4 multiplied by 10-5) titrated by 0.14 M NaOH
(b) 100.0 mL of 0.29 M C2H5NH2 (Kb = 5.6 multiplied by 10-4) titrated by 0.58 M HNO3
(c) 100.0 mL of 0.28 M HCl titrated by 0.14 M NaOH
I'm pretty unclear as far as how to go about it and would greatly appreciate help. Thanks in advance!
The pH at the half-way point of a monoprotic acid is just pKa. For a monoprotic base (C2H5NH2) it is pKa but remember they give you pKb in the problem so pKa = 14-pKb.
The pH at the equivalence point of a monoprotic acid or monoprotic base is calculated from the hydrolysis of the salt. Use (salt) = C = mols salt/L soln.
For the acid the anion is hydrolyzed:
........A^- + HOH ==>HA + OH^-
Kb for the A^- = Kw/Ka for the acid = (x)(x)/(C-x) and solve for x = OH^-, then convert to pH.
For the C2H5NH2 it is the salt C2H4NH3 that is hydrolyzed. I'll call that BNH3^+. Find (BNH3^+) at the equivalence point following the same procedure for the weak acid above.
........BNH3^+ + H2O ==> H3O^+ + BNH2
Ka for BNH3^+ = (Kw/Kb for BNH2) = (x)(x)/(C-x) and solve for x = (H3O^+) and convert to pH.
For the NaOH/HCl (strong base/strong acid),
mols acic to start - mols base added = mols HCl remaining. That gives H^+ when corrected for dilution and pH from that.
The equivalence pint is the hydrolysis of the salt but for SA/SB titrations neither the anion nor the cation are hydrolyzed so the pH = 7.0.
Post your work if you get stuck.
So for the first portion I came to Kb=1.5x10^-10
x^2-(2.2x10^-11)x - 1.56x10^-10=0
solving with the quadratic equation I came to x= 1.25x10^-5 (I'm disregarding i)
So -log(1.25x10^-5)= 4.9
But somewhere I've done something wrong as the homework program I'm using is saying neither 4.9 or 9.1 are correct.
The concn of the salt is not 0.14. Note my instructions were the C = mols/L. At the equivalence point you have (let's call the acid HA)
HA + NaOH ==> NaA + H2O
You have 100 mL x 0.14M acid = 0.014 mols.
You have added 100 mL of 0.14M NaOH so
(salt) = 0.14mol/0.200 L = 0.07M.
You should have ended up with
Kb = 1.56E-10 = (x)(x)/(0.07-x)
If we assume 0.07-x = 0.07, then
x^2 = 1.09E-11 and
x = 3.31E-6 = (OH^-). First we check to see that the assumption is ok. Since 0.07-3.31E-6 = essentially 0.07 the assumption is ok and we need not solve a quadratic.
If OH^- = 3.31E-6 then pOH = 5.48 so pH = 8.52
Alright, I was able to figure them all out. Thank you!
Hi, I need help with a chemistry question.
Question: Iron ore is treated with carbon monoxide to extract and purify the iron.
a) Calculate the minimum mass of carbon monoxide that must be ordered by a refining company for every metric tonne of iron ore that is processed.