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
halfway point
equivalence point
(b) 100.0 mL of 0.29 M C2H5NH2 (Kb = 5.6 multiplied by 10-4) titrated by 0.58 M HNO3
halfway point
equivalence point
(c) 100.0 mL of 0.28 M HCl titrated by 0.14 M NaOH
halfway point
equivalence point
I'm pretty unclear as far as how to go about it and would greatly appreciate help. Thanks in advance!
So for the first portion I came to Kb=1.5x10^-10
then
x^2/.14-x= 1.56x10^-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.
To calculate the pH at the halfway point and the equivalence point in a titration, you need to know the concentrations and dissociation constants of the acid and base involved.
(a) 100.0 mL of 0.14 M HC7H5O2 (Ka= 6.4 multiplied by 10-5) titrated by 0.14 M NaOH:
1. At the halfway point, the acid and base are present in equal moles. This means that half of the acid has been neutralized by the base. To calculate the pH, we need to find the concentration of the remaining acid and the concentration of the conjugate base formed.
Half of the initial 0.14 M HC7H5O2 has reacted, so the concentration of HC7H5O2 remaining is 0.14 M / 2 = 0.07 M. The concentration of the conjugate base (C7H5O2-) formed is also 0.07 M.
To calculate the pH, we can use the Henderson-Hasselbalch equation:
pH = pKa + log10([A-]/[HA])
pKa for HC7H5O2 is given as 6.4 multiplied by 10^-5. Plugging in the values, we get:
pH = -log10(6.4 multiplied by 10^-5) + log10(0.07/0.07)
pH = 4.19
Therefore, at the halfway point, the pH is 4.19.
2. At the equivalence point, all of the acid has been neutralized by the base. This means that the solution only contains the conjugate base formed from the acid.
Since HC7H5O2 is a weak acid, its conjugate base C7H5O2- will undergo hydrolysis. The hydrolysis of C7H5O2- will result in basic conditions.
To determine the pH at the equivalence point, we can calculate the pOH using the concentration of the conjugate base:
pOH = -log10([C7H5O2-])
pOH = -log10(0.14)
Since pOH + pH = 14, we can subtract the pOH from 14 to get the pH:
pH = 14 - pOH
pH = 14 - (-log10(0.14))
pH = 10.85
Therefore, at the equivalence point, the pH is 10.85.
(b) 100.0 mL of 0.29 M C2H5NH2 (Kb = 5.6 multiplied by 10^-4) titrated by 0.58 M HNO3:
1. At the halfway point, half of the base (C2H5NH2) has been neutralized by the acid. The concentration of C2H5NH2 remaining is 0.29 M / 2 = 0.145 M. The concentration of the conjugate acid (C2H5NH3+) formed is also 0.145 M.
Since C2H5NH2 is a weak base, we can use the Henderson-Hasselbalch equation to calculate the pH:
pH = pKb + log10([BH+]/[B])
pKb for C2H5NH2 is given as 5.6 multiplied by 10^-4. Plugging in the values, we get:
pH = -log10(5.6 multiplied by 10^-4) + log10(0.145/0.145)
pH = 9.92
Therefore, at the halfway point, the pH is 9.92.
2. At the equivalence point, all of the base has been neutralized by the acid. This means that the solution only contains the conjugate acid formed from the base.
Since C2H5NH2 is a weak base, its conjugate acid C2H5NH3+ will undergo hydrolysis. The hydrolysis of C2H5NH3+ will result in acidic conditions.
To determine the pH at the equivalence point, we can calculate the pH using the concentration of the conjugate acid:
pH = -log10([C2H5NH3+])
pH = -log10(0.29)
Therefore, at the equivalence point, the pH is 0.54.
(c) 100.0 mL of 0.28 M HCl titrated by 0.14 M NaOH:
Since HCl is a strong acid and NaOH is a strong base, this is a neutralization reaction. At the equivalence point, the moles of HCl and NaOH are equal, resulting in a neutral solution.
Therefore, at the halfway point and at the equivalence point, the pH is 7 (neutral).
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^-
initial.C.............0....0
change..-x............x.....x
equil...C-x...........x.....x
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
initial...C................0.......0
change....-x...............x.......x
equil....C-x...............x.......x
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.