Question

Record your data in the Data Table The data is written in red letters on the screen starting at 1:06 minutes. Sample Data Sodium Chloride (NaCl) (NaBr) a. Starting Mass (8) 23.10 g 75.9g Sodium Nitrate (NaNO3) 80.0 32.18 g 74.9 79.5 b. Ending Mass after Saturation (8) c. Mass Used (8) (b-a) d. Formula Mass 32.18g - 23.10g = 9.08g 75.9-74.9=1.0g 80-79.5=1.3G Na=22.99 ,n=14,0:16 Na: 22.99 g/mole Br: 70.90g/mole NaBr: 22.99+70.90 =102.89 g/mole 9.08g/0.010L = 908 g/L Na:22.9 CI:35.45 NaCl:58.44 e. Solubility in grams per liter c/ 0.010 liters 1.0g/0.01=100g/ 1.3g/0.01=1.30g/ *note: 10 milliliters =0.10 liters f. Solubility in moles per liter eld 100(g//(58.44)=1.71mol/l (908g/L)/102.89 g/mol = 8.82 moles/liter 130(g/l)/52.99g/mol 2.45mol/L

Answer 1

Solubility product Ksp is the product of concentration of dissolved ions of a compound in solution raised to power their stoichiometric coefficients in the balanced chemical dissolution equation for the compound.

Calculation of Ksp for NaCl

Mass of NaCl used=Starting mass-Ending mass(after saturation)

=75.9 g-74.9 g=1.0 g

Formula mass of NaCl=Relative atomic mass of Na+Relative atomic mass of Cl

=22.99 g/mol+35.45 g/mol=58.44 g/mol

Solubility in g/L

=1.0 g/0.01 L=100 g/L

Solubility in mol/L=solubility in g/L/formula mass of NaCl

=100 g/L/(58.44 g/mol)

=1.71 mol/L

Step 1: Balanced equation for dissolution of NaCl in water

NaCl(s) + Na+(aq) + C1-(aq)

Step 2: Solubility of NaCl in mol/L (as calculated above from data in table in image 1)=s=1.71 mol/L

(s=saturated concentration of NaCl)

Let us construct the ICE chart for this as shown

	NaCl	Na+	Cl​​​​​-
Initial	n/a	0	0
Change	n/a	+1.71 mol/L	+1.71 mol/L
Equilibrium	n/a	1.71 mol/L	1.71 mol/L

Step 3:

Ksp=[Na+][Cl​​​​​-]

= s x s=s​​​​​​2

Step 4:

Ksp

=(1.71 mol/L)(1.71 mol/L)

=2.92 mol/L (3 significant figures)

Calculation of Ksp for NaNO​​​3

Mass of NaNO​​​3 used=Starting mass-Ending mass(after saturation)

=80.0 g-79.5 g=0.5 g

Formula mass of NaNO​​​3=Relative atomic mass of Na+Relative atomic mass of N+3xrelative atomic mass of O

=22.99 g/mol+14 g/mol+3x16 g/mol= 84.99 g/mol

Solubility in g/L

=0.5 g g/0.01 L=50 g/L

Solubility in mol/L=solubility in g/L/formula mass of NaCl

=50 g/L/(84.99 g/mol)

=0.588 mol/L

Step 1: Balanced equation for dissolution of NaNO​​​3 in water

NaNO3(s) + Nat(aq) + NO3(aq)

Step 2: Solubility of NaNO​​​3 in mol/L (as calculated above from data in table in image 1)=s=0.588 mol/L

Where s=saturated concentration of NaNO​​​3

Let us construct the ICE chart for this as shown

	NaNO3	Na+	NO3​​​​-
Initial	n/a	0	0
Change	n/a	+0.588 mol/L	+0.588 mol/L
Equilibrium	n/a	0.588 mol/L	0.588 mol/L

Step 3:

Ksp=[Na+][NO3-]

= s x s=s​​​​​​2

Step 4:

Ksp

=(0.588 mol/L)(0.588 mol/L)

= 0.346 mol/L (3 significant figures)