What is the chloride ion concentration, [CT], in a solution made by adding 20.0

Question

What is the chloride ion concentration, [CT], in a solution made by adding 20.0 ml of water to 80.0 ml of a 0.125 M CaCI_2(aq) solution? Write a balanced net ionic equation for the reaction that occurs when aqueous solutions of sodium chlorate and silver nitrate are mixed. Complete and balance the following reaction: Ba(OH)2(aq) + H_2C_2O_4(aq) rightarrow How many mL of 0.100 M KCI(aq) are required to react completely with 4.325 g of solid Pb(NO_3)_2 to form PbCI_2? (i.e., find the amount so that both reactants are completely consumed in the reaction)

Explanation / Answer

8. The molecular equation is AgNO3(aq) + NaCl(aq) --> AgCl(s) + NaNO3

and the net ionic equation is

Ag+(aq) + NO3-(aq) + Na+(aq) + Cl-(aq) --> AgCl(s) + Na+(aq) + NO3-(aq)

9. Ba(OH)2 + H2C2O4 BaC2O4 + 2 H2O

10. Pb(NO3)2(aq) + 2 KCl(aq) = 2 KNO3(aq) + PbCl2

For complete formation of PbCl2 from the reactants given 1 mole of lead nitrate gives 2 moles of potassium chloride.

Pb(NO3)2 mass is 4.325 g which means 4.325/331.22 moles

Molarity = no of moles / volume in liters = 0.1

For every one mole of lead nitrate 2 moles of potassium chloride is required.

Volume required = 4.325/331.22 x 2 x 1/0.1 x 1000 ml (1000 is used to directly convert L to mL )

= 261.155 mL

11. Molecular equation is KOH + CH3COOH = H2O + CH3COOK

Ionic equation, K+ + OH- + CH3COO- + H+ = K+ + CH3COO- + H2O

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