What is the pH at the equivalence point in the
titration of 10.0 mL of 0.6 M HZ with 0.200
M NaOH? Ka = 7.0 × 10−6
for HZ.
The pH will be that of the salt due to hydrolysis.
10.0 mL x 0.6M = ?mL x 0.200 M
?mL = 0.6 x 10/0.2 = 30 mL; therefore, the final volume will be 40 mL for the 6 millimoles for (NaZ) = 6/40 = 0.15M
.............Z + H2O ==> HZ + OH^-
I...........0.15..........0....0
C............-x............x.....x
E..........0.15-x........x.......x
Kb(for Z^-) = (Kw/Ka for Hz) = (HZ)(OH^-)/(Z^-)
Substitute and solve for x = OH^- then convert to pH.
To find the pH at the equivalence point in the titration of 10.0 mL of 0.6 M HZ with 0.200 M NaOH, we need to determine the amount of NaOH required to neutralize the HZ.
First, we need to calculate the number of moles of HZ present in 10.0 mL of 0.6 M HZ:
moles of HZ = volume (in L) x concentration (in mol/L)
moles of HZ = 10.0 mL x (1 L / 1000 mL) x 0.6 mol/L
moles of HZ = 0.006 mol
Since the balanced equation for the reaction between HZ and NaOH is:
HZ + NaOH -> NaZ + H2O
we can see that the stoichiometric ratio is 1:1. This means that 1 mole of HZ reacts with 1 mole of NaOH.
Therefore, the amount of NaOH required to neutralize 0.006 moles of HZ is also 0.006 moles.
Now, we can calculate the volume of 0.200 M NaOH required to neutralize 0.006 moles of HZ:
volume (in L) = moles / concentration (in mol/L)
volume (in L) = 0.006 mol / 0.200 mol/L
volume (in L) = 0.030 L
Since the initial volume of HZ is 10.0 mL (or 0.010 L) and we added 0.030 L of NaOH, the total volume at the equivalence point is 0.040 L.
Next, we need to determine the concentration of the resulting solution at the equivalence point. Since the stoichiometric ratio is 1:1, the moles of NaOH at the equivalence point is also 0.006 mol.
The total volume of the resulting solution is 0.040 L, so the concentration of NaOH at the equivalence point is:
concentration (in mol/L) = moles / volume (in L)
concentration (in mol/L) = 0.006 mol / 0.040 L
concentration (in mol/L) = 0.150 M
Now, we can calculate the pOH at the equivalence point:
pOH = -log10(concentration of OH- ions)
pOH = -log10(0.150)
pOH = 0.823
Finally, we can find the pH at the equivalence point by using the equation:
pH + pOH = 14
pH + 0.823 = 14
pH = 14 - 0.823
pH = 13.177
Therefore, the pH at the equivalence point in the titration is approximately 13.177.
To determine the pH at the equivalence point in this titration, we need to understand the concept of an equivalence point and the chemistry behind it.
In a titration, an acid and a base are reacted together. The equivalence point occurs when the number of moles of acid is equal to the number of moles of base added. At this point, all the acid has been neutralized by the base, resulting in a solution that contains only the conjugate base of the acid.
In this specific titration, we are adding NaOH (base) to HZ (acid). HZ has a dissociation constant, Ka, of 7.0 × 10^(-6).
To find the pH at the equivalence point, we can use the Henderson-Hasselbalch equation, which relates the pH of a solution to the concentration of the acid and its conjugate base:
pH = pKa + log([A-]/[HA])
Where [A-] is the concentration of the conjugate base and [HA] is the concentration of the acid.
At the equivalence point, all of the HZ has been neutralized, resulting in a solution that contains only the conjugate base, A-. Therefore, the concentration of HZ is zero, and the concentration of the conjugate base is equal to the initial concentration of HZ.
In this case, the initial concentration of HZ is 0.6 M. Therefore, the concentration of A- at the equivalence point also becomes 0.6 M.
Now, we need the pKa value to calculate the pH. Given that Ka = 7.0 × 10^-6, we can determine pKa using the following formula:
pKa = -log10(Ka)
pKa = -log10(7.0 × 10^-6)
pKa ≈ 5.154
Substituting the values into the Henderson-Hasselbalch equation, we get:
pH = 5.154 + log(0.6/0)
However, the logarithm of zero is undefined. Since the concentration of HA is zero at the equivalence point, we need to use the limit concept to solve this equation.
lim [HA]→0 log(0.6/[HA])
As the concentration of HA approaches zero and the concentration of A- becomes significant, the logarithm term approaches negative infinity. Therefore, at the equivalence point, the pH would be very close to infinity or extremely basic.
So, in conclusion, at the equivalence point in the titration of 10.0 mL of 0.6 M HZ with 0.200 M NaOH, the pH would be very close to infinity or extremely basic.