Calculate the pH of each of the following aqueous solutions. (Enter your answers to two decimal places.) (a) 10.0 mL deionized water WebAssign will check your answer for the correct number of significant figures. 2.72 Incorrect: Your answer is incorrect. (b) 10.0 mL deionized water plus 5.0 mL of 0.10 M NaOH WebAssign will check your answer for the correct number of significant figures. (c) 10.0 mL deionized water plus 10.0 mL of 0.10 M NaOH WebAssign will check your answer for the correct number of significant figures. (d) 10.0 mL deionized water plus 15.0 mL of 0.10 M NaOH WebAssign will check your answer for the correct number of significant figures.
Answer:
a. pH = 7.0
b. pH = 12.52
c. pH = 12.70
d. pH = 12.78
Explanation:
a. Deionized water has the [H⁺] of pure water = 1x10⁻⁷ (Kw = 1x10⁻¹⁴ = [H⁺][OH⁻] - [H⁺] = [OH⁻ -)
pH = -log[H⁺] = 7
b. Moles NaOH = 5x10⁻³L * (0.10mol / L) = 5x10⁻⁴moles OH⁻ / 0.015L = 0.0333M = [OH⁻]
-Total volume = 10mL+5mL = 15mL = 0.015L
pOH = -log[OH⁻] = 1.48
pH = 14-pOH
pH = 12.52
c. Moles NaOH = 0.010L * (0.10mol / L) = 1x10⁻³moles OH⁻ / 0.020L = 0.0500M = [OH⁻]
-Total volume = 10mL+10mL = 20mL = 0.020L
pOH = -log[OH⁻] = 1.30
pH = 14-pOH
pH = 12.70
d. Moles NaOH = 0.015L * (0.10mol / L) = 1.5x10⁻³moles OH⁻ / 0.025L = 0.060M = [OH⁻]
-Total volume = 10mL+15mL = 25mL = 0.025L
pOH = -log[OH⁻] = 1.22
pH = 14-pOH
pH = 12.78
Which change represents an oxidation reaction?
An empty steel container is filled with 2.0 atm of H₂ and 1.0 atm of F₂. The system is allowed to reach equilibrium according to the reaction below. If Kp = 0.45 for this reaction, what is the equilibrium partial pressure of HF?
The equilibrium partial pressure of HF is 0.55 atm.
The equation of the reaction is;
H2(g) + F2(g) ⇄ 2HF
I 2 1 0
C -x -x +x
E 2 - x 1 - x x
We know that;
pH2 = 2.0 atm
PF2 = 1.0 atm
pHF = ??
Kp = 0.45
So;
Kp = (pHF)^2/pH2. pF2
0.45 = x^2/(2 - x) (1 - x)
0.45 = x^2/x^2 - 3x + 2
0.45(x^2 - 3x + 2) = x^2
0.45x^2 - 1.35x + 0.9 = x^2
0.55 x^2 + 1.35x - 0.9 = 0
x = 0.55 atm
Learn more about equilibrium: https://brainly.com/question/3980297
Lithium and nitrogen react to produce lithium nitride:
6Li (s) + N2 (g) = 2Li3N (s)
How many moles of lithium nitride are produced when 0.400 mol lithium are used in this reaction?
a. 1.28 mol
b. 0.133 mol
c. 0.800 mol
d. 0.0667 mol
Answer:
B : 0.133 M
Explanation:
moles Li3N3 = 0.4 mol Li x (2 moles Li3N/6 moles Li) = 0.133 M