1) A saturated solution of the slightly soluble salt MA3 is 1.4 ×10^–3 M in MA3. What is its Ksp value?
2) A 20.0 mL sample of NaIO3 was treated with an excess of KI and HCl and the iodine produced titrated with 0.05125 M Na2S2O3. 19.35 mL of Na2S2O3 solution were required to reach a starch‐iodine endpoint. What was the concentration of the NaIO3?
To answer these questions, we need to understand the concept of equilibrium and the solubility product constant (Ksp). Let's break down each question and explain how to solve them.
1) A saturated solution of the slightly soluble salt MA3 is 1.4 ×10^–3 M in MA3. What is its Ksp value?
The solubility product constant (Ksp) is a value that represents the equilibrium concentration of the ions in a saturated solution of a sparingly soluble salt. In this case, we have a saturated solution of the salt MA3 with a concentration of 1.4 ×10^–3 M.
The salt MA3 dissociates into M^+ and 3A^- ions according to the equilibrium equation: MA3 ⇌ M^+ + 3A^-
To find the Ksp value, we need to determine the equilibrium concentrations of the ions. Since every 1 mole of MA3 dissociates to 1 mole of M^+ and 3 moles of A^-, the equilibrium concentrations will be:
[M^+] = 1.4 ×10^–3 M
[A^-] = 3 × 1.4 × 10^–3 M = 4.2 × 10^–3 M
Now, we can plug these values into the equilibrium expression for Ksp:
Ksp = [M^+][A^-]^3
Ksp = (1.4 ×10^–3 M)(4.2 × 10^–3 M)^3 = 1.327 × 10^–11
Therefore, the Ksp value for the slightly soluble salt MA3 is 1.327 × 10^–11.
2) A 20.0 mL sample of NaIO3 was treated with an excess of KI and HCl, and the iodine produced was titrated with 0.05125 M Na2S2O3. 19.35 mL of Na2S2O3 solution were required to reach a starch‐iodine endpoint. What was the concentration of the NaIO3?
In this question, we need to determine the concentration of NaIO3 given the volume of Na2S2O3 solution required to react with the iodine produced.
The balanced chemical equation for the reaction between NaIO3 and KI in the presence of HCl is:
5NaIO3 + 5KI + 6HCl → 3I2 + 5KCl + 3H2O + 5NaCl
From the equation, we can see that 5 moles of NaIO3 react to produce 3 moles of I2. Therefore, there is a 3:5 molar ratio between I2 and NaIO3.
The volume of Na2S2O3 required to reach the starch-iodine endpoint represents the amount of iodine (I2) produced. So we need to determine the amount of iodine produced, which is in moles, using the equation:
moles of I2 = (0.05125 M Na2S2O3) x (19.35 mL) x (1 L/1000 mL)
Next, we need to convert the moles of iodine to moles of NaIO3 using the 3:5 molar ratio:
moles of NaIO3 = (3/5) x moles of I2
Finally, we can calculate the concentration of NaIO3 in the 20.0 mL sample:
concentration of NaIO3 = moles of NaIO3 / (20.0 mL / 1000)
By following these steps, you can determine the concentration of NaIO3 in the given sample.