Question

1)Another metal phosphate is aluminum phosphate. It will behave similar to calcium phosphate in an acid solution. What is the net ionic equation including phases for AlPO4(s) dissolving in H3O+(aq)?

Express your answer as a net ionic equation.

2)A 75.0-mL volume of 0.200 M NH3 (Kb=1.8×10−5) is titrated with 0.500 M HNO3. Calculate the pH after the addition of 27.0 mL of HNO3.

3)A 52.0-mL volume of 0.35 M CH3COOH (Ka=1.8×10−5) is titrated with 0.40 M NaOH. Calculate the pH after the addition of 25.0 mL of NaOH.

4) What is the solubility of M(OH)2 in a 0.202 M solution of M(NO3)2?

5)

As a chemist for an agricultural products company, you have just developed a new herbicide,"Herbigon," that you think has the potential to kill weeds effectively. A sparingly soluble salt, Herbigon is dissolved in 1 M acetic acid for technical reasons having to do with its production. You have determined that the solubility product Ksp of Herbigon is 9.00×10⁻⁶.

Although the formula of this new chemical is a trade secret, it can be revealed that the formula for Herbigon is X-acetate (XCH3COO, where "X" represents the top-secret cation of the salt). It is this cation that kills weeds. Since it is critical to have Herbigon dissolved (it won't kill weeds as a suspension), you are working on adjusting the pH so Herbigon will be soluble at the concentration needed to kill weeds. What pH must the solution have to yield a solution in which the concentration of X+ is 5.50×10⁻³M ? The pKa of acetic acid is 4.76.

Express your answer numerically.

Answer 1

1 ) AlPO4 (s) + H3O+ (aq) ---------------------> Al+3 (aq) + HPO4-2 (aq)

2)

millimoles of NH3 = 75 x 0.2 = 15

millimoles of H+ = 27 x 0.5 = 13.5

NH 3 + H+ -----------------------> NH4+

15        13.5                                0

1.5        0                                   13.5

pOH = pKb + log [NH4+] / NH3]

pOH = 4.74 + log (13.5/1.5)

pOH = 5.69

pH + pOH = 14

pH = 8.31

3)

millimoles of CH3COOH = 52 x 0.35 = 18.2

millimoles of NaOH = 25 x 0.4 = 10

millimoles of CH3COOH reamined = 18.2 - 10 = 8.2

salt formed = 10

pH = pKa + log [CH3COONa]/[CH3COOH]

pH = 4.74 + log (10 /8.2)

pH = 4.83

4)

you missed to give Ksp of M(OH)2