3. A solution was prepared by dissolving 1.564 g of NaCl and 2.765 g of KCl in s

Question

3. A solution was prepared by dissolving 1.564 g of NaCl and 2.765 g of KCl in some water and diluting to exactly 250 mL in a graduated flask. What is the molarity of Na+, K+ and Cl- ions in the solution?

4. Suppose you have a solution that might contain Al3+, Ca2+, Ag+, and Zn2+ ions. Addition of HCl gave a white precipitate. After filtering off the precipitate, potassium sulfate was added to the filtrate resulting in a white precipitate, which was again filtered off. When a large amount of KOH was added to the filtrate, no precipitate was formed. Which cations are present in the solution? I need help with those three questions please.

Explanation / Answer

3)

volume = 250 mL = 0.250 L

number of moles of NaCl = mass / molar mass

= 1.564 / 58.5

= 0.0267 mol

[NaCl] = number of moles of NaCl / volume

= 0.0267 mol / 0.250 L

= 0.107 M

number of moles of KCl = mass / molar mass

= 2.765 / 74.55

= 0.0371 mol

[KCl] = number of moles of KCl / volume

= 0.0371 mol / 0.250 L

= 0.148 M

So,

[Na+] = [NaCl] = 0.107 M

[K+] = [KCl] = 0.148 M

[Cl-] = [NaCl] + [KCl] = 0.107 + 0.148 = 0.255 M

I am allowed to answer only 1 question at a time

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