Answer:
The Ka is 9.11 *10^-8
Explanation:
<u>Step 1: </u>Data given
Moles of HX = 0.365
Volume of the solution = 835.0 mL = 0.835 L
pH of the solution = 3.70
<u>Step 2:</u> Calculate molarity of HX
Molarity HX = moles HX / volume solution
Molarity HX = 0.365 mol / 0.835 L
Molarity HX = 0.437 M
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<u>Step 3:</u> ICE-chart
[H+] = [H3O+] = 10^-3.70 = 1.995 *10^-4
Initial concentration of HX = 0.437 M
Initial concentration of X- and H3O+ = 0M
Since the mole ratio is 1:1; there will react x M
The concentration at the equilibrium is:
[HX] = (0.437 - x)M
[X-] = x M
[H3O+] = 1.995*10^-4 M
Since 0+x = 1.995*10^-4 ⇒ x=1.995*10^-4
[HX] = 0.437 - 1.995*10^-4 ≈ 0.437 M
[X-] = x = 1.995*10^-4 M
<u>Step 4: </u>Calculate Ka
Ka = [X-]*[H3O+] / [HX]
Ka = ((1.995*10^-4)²)/ 0.437
Ka = 9.11 *10^-8
The Ka is 9.11 *10^-8
Answer:
The equilibrium temperature of the coffee is 72.4 °C
Explanation:
Step 1: Data given
Mass of cream = 15.0 grams
Temperature of the cream = 10.0°C
Mass of the coffee = 150.0 grams
Temperature of the coffee = 78.6 °C
C = respective specific heat of the substances( same as water) = 4.184 J/g°C
Step 2: Calculate the equilibrium temperature
m(cream)*C*(T2-T1) = -m(coffee)*c*(T2-T1)
15.0 g* 4.184 J/g°C *(T2 - 10.0°C) = -150.0g *4.184 J/g°C*(T2-78.6°C)
62.76T2 - 627.6 = -627.6T2 + 49329.36
690.36T2 = 49956.96
T2 = 72.4 °C
The equilibrium temperature of the coffee is 72.4 °C
