Chemistry 12th Grade Homework: NaOH, HCl, AgBr, NH3, HCl, Phosphoric Acid

[JoanF]

Homework Statement

1 - An aqueous solution of NaOH (strong base), of concentration 0.10 mol/L was gradually added to a solution of HCl (strong acid), molar concentration of 0.08 mol/L. Select the graphic that provides the concentrations of different species during this addition.

2- AgBr (s) -> Ag+ (aq) + Br- (aq), Ks=7,7*10^-13 and Kf=10^21

2.1 - calc the solubility of AgBr in water.
2.2 - calc the solubility of AgBr in a solution of KCN 0,10 mol/L.

3 - A sample of 10.0 mL of a solution of NH3 (aq) concentration 0.15 mol/dm^3, is titrated with a solution 0.10 mol/dm^3 in HCl (aq).
3.1 - What volume of HCl worn to the point of equivalence?
3.2 - What the pH at equivalence point?
3.3 - What is the pH at the time of titration is added to half the volume of titrant worn to the point of equivalence.?

4 - The following figure (figure number 2) represents the titration of a solution of phosphoric acid with a solution of sodium hydroxide.
As phosphoric acid is an acid polyprotic the titration curve has several points of equivalence and also a number of buffer zones.
4.1 - What are the buffer zones and the species that determine the buffer effect in each of these areas?
4.2 - What relationship exists between the volume of titrant spent by the first equivalence point and the volume of titrant spent between the first and second equivalence point?

Homework Equations

The Attempt at a Solution

1 - I really don't know how to do this :S
2 - 2.1 I got 8.8*10^-7 mol/dm^3
2.2 i think it is 8.8*10^-7 mol/dm^3, too.

3 - I ve just write the eq.: NH3 + HCl -> NH4Cl but I don't know how o solve the problem.

4.1 -
4.2 - they're equal.



Please help me...I've test tomorrow :S :S :S

 

[nate808]

1. The correct graphic for this scenario would be a titration curve, which shows the concentration of different species (in this case, NaOH and HCl) as the volume of titrant (NaOH) is gradually added to the solution of HCl.

2.1. To calculate the solubility of AgBr in water, we can use the solubility product constant (Ks) equation: Ks = [Ag+][Br-]. Since we know the value of Ks (7.7*10^-13) and the concentration of Ag+ and Br- are equal (x), we can set up the equation: 7.7*10^-13 = x^2. Solving for x, we get a solubility of 8.8*10^-7 mol/L.

2.2. To calculate the solubility of AgBr in a solution of KCN, we need to take into account the formation constant (Kf) for the complex ion Ag(CN)2-. This complex ion is formed by the reaction Ag+ + 2CN- -> Ag(CN)2-. Since we have a concentration of 0.10 mol/L for KCN, we can calculate the concentration of CN- (0.20 mol/L). Using the equilibrium constant equation, Kf = [Ag(CN)2-]/[Ag+][CN-]^2, we can set up the following equation: 10^21 = x/(0.20)^2. Solving for x, we get a solubility of 1.0*10^-8 mol/L.

3.1. To determine the volume of HCl needed to reach the equivalence point, we can use the equation M1V1 = M2V2, where M1 is the concentration of NH3 (0.15 mol/dm^3), V1 is the initial volume of NH3 (10.0 mL), M2 is the concentration of HCl (0.10 mol/dm^3), and V2 is the final volume of HCl at the equivalence point. Rearranging the equation, we get V2 = (M1V1)/M2 = (0.15*10.0)/0.10 = 15.0 mL.

3.2. At the equivalence point, all the NH3 has been neutralized by the HCl, resulting in a solution