Calculate the pH at the equivalence point in titrating a 0.120 M solution of NaHCrO4 (sodium hydrogen chromate) with 7.0×10−2 NaOH.
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Calculate the pH at the equivalence point in titrating 0.120 M solutions of each of the following with 9.0×10−2 MNaOH
To calculate the pH at the equivalence point, we need to determine the concentration of the hydronium ion (H3O+) or hydroxide ion (OH-) at that point. The equivalence point occurs when the moles of the two reacting species are present in stoichiometric proportions.
In this case, we have a titration of a weak acid (NaHCrO4) with a strong base (NaOH). Therefore, at the equivalence point, the weak acid will be completely neutralized by the strong base.
To find the pH at the equivalence point, we need to first determine the balanced chemical equation for the reaction between NaHCrO4 and NaOH.
NaHCrO4 + NaOH → Na2CrO4 + H2O
From the balanced equation, we can see that the stoichiometric ratio between NaHCrO4 and NaOH is 1:1. This means that when all the NaHCrO4 has reacted with the NaOH, the number of moles of NaOH will be equal to the number of moles of NaHCrO4 initially present.
To determine the moles of NaOH used at the equivalence point, we can use the molarity of the NaOH solution and its volume.
Moles of NaOH = Molarity * Volume
Given:
Molarity of NaOH solution (C_NaOH) = 7.0×10^−2 M
Volume of NaOH solution (V_NaOH) = unknown
Next, we need to determine the initial moles of NaHCrO4 present in the solution. This can be calculated using the molarity of NaHCrO4 and its volume.
Moles of NaHCrO4 = Molarity * Volume
Given:
Molarity of NaHCrO4 solution (C_NaHCrO4) = 0.120 M
Volume of NaHCrO4 solution (V_NaHCrO4) = unknown
Since both NaHCrO4 and NaOH have a 1:1 stoichiometric ratio, the moles of NaHCrO4 will be equal to the moles of NaOH at the equivalence point. Therefore,
Moles of NaHCrO4 initial = Moles of NaOH at equivalence point
Set up the equation:
C_NaHCrO4 * V_NaHCrO4 = C_NaOH * V_NaOH
We can solve this equation to find the volume of NaOH used at the equivalence point (V_NaOH).
Once we know the volume of NaOH used at the equivalence point, we can assume that the volume of NaHCrO4 solution and NaOH solution are added together. Therefore, the final volume is the sum of the volumes of NaHCrO4 and NaOH.
The final volume of the solution (V_final) = V_NaHCrO4 + V_NaOH
Now, to calculate the concentration of the hydronium ion (H3O+) at the equivalence point, we can use the fact that NaHCrO4 is a weak acid. The dissociation of NaHCrO4 can be represented by the following equation:
NaHCrO4 + H2O ↔ HCrO4- + Na+ + OH-
From the balanced equation, we can see that the concentration of OH- will be equal to the concentration of NaHCrO4 at the equivalence point.
Concentration of OH- = C_NaHCrO4
Since the concentration of H3O+ and OH- are related by the autoprotolysis constant of water (Kw), we can use the equation:
[H3O+] * [OH-] = Kw
[H3O+] = Kw / [OH-]
where Kw = 1.0 × 10^-14 (at 25°C).
Substituting the value of [OH-] with the concentration of NaHCrO4 at the equivalence point, we can calculate the concentration of H3O+.
[H3O+] = (1.0 × 10^-14) / C_NaHCrO4
Finally, to find the pH, we can take the negative logarithm of the H3O+ concentration using the formula:
pH = -log[H3O+]
Plug in the value of [H3O+] to calculate the pH at the equivalence point.