a .4000 M solution of nitric acid is used to titrate 50.00 mL of .237 M barium hydroxide (Assume that volumes are additive).
a) Write a balanced net ionic equation for the reaction that takes place during titration.
b) what are the species present at the equivalence point?
c) what volume nitric acid is required to reach the equivalence point?
d) what is the pH of the solution before any nitric acid is added?
e) what is the pH of the solution halfway to the equivalence point?
f) what is the pH of the solution at the equivalence point?
I said the net ionic equation was:
OH- + H+ --> H2O
And the species present are: Ba+2 and NO3-
I thought because, at the equivalence point only spectator ions are present, the equivalence point is 7. I was considering, for c), to say 7= pKa + log (base/acid), but since HNO3 is a strong acid, it wont have a Ka correct? So how do I approach this and letter d)?
a) is ok.
b) what you have answered is good; there will be a small amount of H^+ and OH^- there from the ionization of water but it is almost miniscule. (H^+)=(OH^-)= 1E-7 M. I don't know if that is neglibible in terms of your problem or not since I don't know the history of the class.
c)volume of HNO3 required to reach the equivalence point.
The equation is
2HNO3 + Ba(OH)2 ==> 2H2O + Ba(NO3)2
mols Ba(OH)2 = 0.05 L x 0.237 M = 0.01185.
mols HNO3 required = 0.01185 x (2 mols HNO3/1 mol Ba(OH)2) = 0.01185 x 2 = 0.02370.
L x M = mols; therefore, L = mols/M = 0.02370 mols HNO3/0.4000 = ?? Liters.
d)pH of Ba(OH)2 :
Ba(OH)2(aq) ==> Ba^+2 + 2OH^-
M Ba(OH)2 = 0.237 M = 0.237 mols/L. Therefore, (OH^-) = 2 x 0.237 = ??
pOH = - log (OH^-)=??
pH + pOH = pKw = 14
Solve for pH.
e). Halfway to the equivalence point:
You know the molarity of the Ba(OH)2 at the beginning. Twice that will be the (OH^-) at the beginning and that answer is the same as from part d. Then half that will be the (OH^-) when it is half gone. Calculate pOH and pH from that.
f). 7 is correct.
At the equivalence point, the moles of HNO3 added will equal the moles of Ba(OH)2 initially present. Therefore, the (OH^-) = 0 and the (H^+) = 0. The pH = pOH = 7.
a) The balanced net ionic equation for the reaction during titration is:
2H+ + 2OH- --> 2H2O
b) At the equivalence point, the species present are Ba2+ and NO3-. There will also be a small amount of H+ and OH- from the ionization of water, but it is negligible.
c) To determine the volume of nitric acid required to reach the equivalence point, we need to use stoichiometry. From the balanced equation, we can see that 2 moles of HNO3 react with 1 mole of Ba(OH)2.
First, calculate the moles of Ba(OH)2:
moles of Ba(OH)2 = volume of Ba(OH)2 x molarity of Ba(OH)2
= 50.00 mL x 0.237 M
= 0.01185 moles
Since the ratio is 2 moles of HNO3 to 1 mole of Ba(OH)2, the moles of HNO3 required will be:
moles of HNO3 = 0.01185 moles Ba(OH)2 x (2 moles HNO3 / 1 mole Ba(OH)2)
= 0.0237 moles HNO3
Finally, calculate the volume of nitric acid required:
volume of HNO3 = moles of HNO3 / molarity of HNO3
= 0.0237 moles / 0.4000 M
= 0.05925 L
= 59.25 mL
Therefore, 59.25 mL of nitric acid is required to reach the equivalence point.
d) To determine the pH of the solution before any nitric acid is added, we need to calculate the concentration of OH- from Ba(OH)2.
Concentration of Ba(OH)2 = 0.237 M
Since Ba(OH)2 is a strong base, it will dissociate completely, and the concentration of OH- will be twice the concentration of Ba(OH)2:
Concentration of OH- = 2 * 0.237 M = 0.474 M
Now, calculate the pOH:
pOH = -log10(0.474) = 0.324
Since pH + pOH = 14, we can calculate the pH:
pH = 14 - 0.324 = 13.676
Therefore, the pH of the solution before any nitric acid is added is 13.676.
e) Halfway to the equivalence point, half of the moles of Ba(OH)2 will have reacted. So, the moles of Ba(OH)2 remaining will be:
moles of Ba(OH)2 remaining = moles of Ba(OH)2 initial / 2
= 0.01185 moles / 2
= 0.005925 moles
The concentration of Ba(OH)2 remaining will be:
concentration of Ba(OH)2 remaining = moles of Ba(OH)2 remaining / volume of solution
= 0.005925 moles / 0.0500 L
= 0.1185 M
The concentration of OH- remaining will be twice the concentration of Ba(OH)2 remaining:
Concentration of OH- = 2 * 0.1185 M = 0.237 M
Now, calculate the pOH:
pOH = -log10(0.237) = 0.625
Since pH + pOH = 14, we can calculate the pH:
pH = 14 - 0.625 = 13.375
Therefore, the pH of the solution halfway to the equivalence point is 13.375.
f) At the equivalence point, all the moles of Ba(OH)2 will have reacted, producing equal moles of H+ and OH-. Therefore, the concentration of H+ and OH- will be equal, resulting in a pH of 7 (neutral).
a) The balanced net ionic equation for the reaction that takes place during titration is as follows:
2HNO3 + Ba(OH)2 -> 2H2O + Ba(NO3)2
b) At the equivalence point, the species present are Ba+2 and NO3-. There will also be a small amount of H+ and OH- ions from the ionization of water, but their concentrations are negligible in this context.
c) To calculate the volume of nitric acid required to reach the equivalence point, we need to determine the number of moles of Ba(OH)2. First, calculate the moles of Ba(OH)2:
moles of Ba(OH)2 = volume (in L) x concentration (in M) = 0.05 L x 0.237 M = 0.01185 mol
Since the stoichiometric ratio of HNO3 to Ba(OH)2 is 2:1, the moles of HNO3 required are:
moles of HNO3 = 0.01185 mol x (2 mol HNO3 / 1 mol Ba(OH)2) = 0.02370 mol
Finally, calculate the volume of nitric acid required:
volume of HNO3 = moles of HNO3 / concentration of HNO3 = 0.02370 mol / 0.4000 M = 0.0593 L (or 59.3 mL)
d) Before any nitric acid is added, the solution consists of only barium hydroxide (Ba(OH)2) with a concentration of 0.237 M. To find the pH of this solution, we need to calculate the pOH and then convert it to pH.
Since Ba(OH)2 dissociates into Ba+2 and 2OH-, the concentration of hydroxide ions (OH-) is 2 times the concentration of Ba(OH)2:
[OH-] = 2 x 0.237 M = 0.474 M
Now, calculate the pOH:
pOH = -log[OH-] = -log(0.474) ≈ 0.325
Since pH + pOH = 14, we can find the pH by subtracting the pOH from 14:
pH = 14 - pOH = 14 - 0.325 ≈ 13.675
e) Halfway to the equivalence point, we still have some barium hydroxide (Ba(OH)2) remaining. The concentration of Ba(OH)2 at the beginning is 0.237 M, so twice that will be the concentration of hydroxide ions (OH-) at the beginning, which is the same as in part d: 0.474 M. Half of that concentration will be the concentration of hydroxide ions (OH-) halfway to the equivalence point:
[OH-] = 0.474 M / 2 = 0.237 M
Now, calculate the pOH and pH using the same method as in part d.
f) The pH of the solution at the equivalence point is 7. At the equivalence point, all the barium hydroxide has reacted with the nitric acid, resulting in a neutral solution with equal concentrations of H+ and OH- ions, leading to a pH of 7.