If 2.00 mL of0.400 M NaOH are added to 1.000 L of0.700 M CaCI2, what is the valu

Question

If 2.00 mL of0.400 M NaOH are added to 1.000 L of0.700 M CaCI2, what is the value of the reaction quotient and will precipitation occur? Which of the following represents the solubility product expression for the ionic compound A_xB_y(s)? K_sp = [A^y+]^y [B^x-]^x K_sp = [A^y+] [B^x-] K_sp = [A^y+]^x [B^x-]^y K_sp = [A^y+]^x [B^x-]^xy/[a^xB^y] Calculate the concentration of ions in the following saturated solutions: [I^-] in AgI solution with [Ag^+] = 5.9 Times 10^-8 M [Al^3+] in Al (OH)_3 solution with [OH^-] = 2.9 Times 10^-9 M From the solubility data given, calculate the solubility products for the following compounds: SrF_2. 7.3 Times 10^-2 g/L Ag_3PO_4. 6.7 Times 10^-3 g/L

Explanation / Answer

Answer for Q5:

Given reagents - 2 ml of 0.4M NaOH, 1 L of 0.7M CaCl2

For the given reagents precipitation reaction can be written as

2NaOH+CaCl2 ------> 2NaCl+Ca(OH)2

For the simplification above reaction can be assumed as A+B -----> C+D

[C]c[D]d

The reaction quotient is given as Q= -------------- where a,b,c,d are mole concentrations of each components.

[A]a[B]b

From the given reaction , reaction qotient can be calculated as

[NaCl]2[Ca(OH)2]1   3

Q= ---------------------------- = -------------- = 1

[NaOH]2[CaCl2]1 3   

The reaction quotient for given reagents is 1.

Answer for Q8:

a). When SrF2 Desolves it dissociates like the following reaction.

SrF2 Sr2+ + 2F¯

The Solubility (K) is written as :

K = [Sr2+] [F¯]2

There is a 1:1 molar ratio between SrF2 and Sr2+, BUT there is a 1:2 molar ratio between SrF2 and F¯. This means that, when 7.3 x 10¯2 mole per liter of SrF2 dissolves, it produces 7.3 x 10¯2 mole per liter of Sr2+ and it produces 14.6 x 10¯2 mole per liter of F¯ in solution.

Putting the values into the K expression, we obtain:

K = (7.3 x 10¯2) (14.6 x 10¯2)2 = 3.92 x 10¯11 = 0.001556

Therefore K=0.001556

b).When Ag3PO4 Desolves it dissociates like the following reaction.

The K expression is:

K = [Ag2+]3 [PO43¯]2

We know the following:

These is a 3:1 ratio between the concentration of the Silver ion and the molar solubility of the Silver phosphate.

There is a 2:1 ratio between the concentation of the phosphate ion and the molar solubility of the magnesium phosphate.

Therefore:

K = (6.7 x 10¯3)3 (13.4 x 10¯3)2

K = 8.06 x 10¯8.

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