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⁻]
<em>-Total volume = 10mL+5mL = 15mL = 0.015L</em>
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⁻]
<em>-Total volume = 10mL+10mL = 20mL = 0.020L</em>
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⁻]
<em>-Total volume = 10mL+15mL = 25mL = 0.025L</em>
pOH = -log[OH⁻] = 1.22
pH = 14-pOH
pH = 12.78
The answer is <span>572.1235 grams of co2</span>
Hard question thx for the points
Collect the salt water and put it in something where it can boil under fire or any other heat source capable to bring the water to boiling temperature. You can consume the collected water safely , but wait for its temperature to fall, except if you need it to be hot.