Question

Given 855cm^3 of a 0.00330M silver nitrate is mixed with 424cm^3 of a .0110M potassium chromate find,

a) balanced equation for both substances when placed in water

b) equation for the net ionic reaction

c) Ksp expression

d) concentration of all ions in solution after mixing but before any reaction takes place

Answer 1

Given: Volume of silver nitrate = 855 cm³ = 855 mL = 0.855 L

Concentration of Silver nitrate = 0.00330 M

Volume of potassium chromate = 424 cm³ = 424 mL = 0.424 L , molarity = 0.0110 M

1. Balanced equation

2AgNO3 (aq) + K2CrO4 (aq)---- > Ag2CrO4 (s)+2 KNO3(aq)

1. Equation for net reaction :

To write net ionic equation we split each aqueous terms into ions.

2AgNO3 (aq) + K2CrO4 (aq)---- > Ag2CrO4 (s)+2 KNO3(aq)

2Ag+ (aq) + 2 NO3- (aq) + 2 K+ (aq) + CrO4 2- (aq)--- > Ag2CrO4 (s) + 2 K+ 2 NO3-

We cancel the spectator ions

2Ag+ (aq) + CrO4 2- (aq)--- > Ag2CrO4 (s)

1. Ksp expression for Ag2CrO4 (s)

Ksp = [Ag]²[CrO42-]

1. Lets calculate concentration of all ions:

Lets calculate number of moles of AgNO3 =Concentration * Volume in L

= 0.00330 M * 0.855L= 0.002822 mol

Number of moles ofAg+ = NO3- = moles of AgNO3

Therefore moles of Ag+ = 0.002822 M

And moles of of NO3-= 0.002822 M

[Ag+] 0.002822 mol / ( 0.855+0.424 ) L

= 0.0022 M

[NO3 - ]= 0.0022 M

To calculate concentration of chromate ion and pottasium ions we calcualte moels of potassium chromate

Mol K2CrO4 = 0.00110 M * 0.424 L

=0.0004664 mol

Mol K+ = 2 * 0.000464 = 0.000933 mol

mol CrO4 2- = mol K2CrO4 = 0.000464 mol

[K+] = 0.000933 mol / ( 0.855+0.424) L = 0.000729 M

[CrO42- ]= 0.000464 / ( 0.855 + 0.424) =0.0003647 M