Since you didn't give the actual volume (or any of the experimental values) I can only tell you how to do it. Do the calculation using the real (determined) volume of the flask. Then, re-do the calculation with v = 125ml. Take the two values and calculate % error; m = measured vol; g = guessed vol.
<span>[mW (m) - mW (g)]/mW (m) x 100% </span>
<span>(they want % error so, if it is negative, just get rid of the sign) </span>
Nitrogen I believe . I need 20 characters.
Answer:
[OH⁻] = 3.34x10⁻³M; Percent ionization = 0.54%; pH = 11.52
Explanation:
Kb of the reaction:
NH3 + H2O(l) ⇄ NH4+ + OH-
Is:
Kb = 1.8x10⁻⁵ = [NH₄⁺] [OH⁻] / [NH₃]
<em>As all NH₄⁺ and OH⁻ comes from the same source we can write: </em>
<em>[NH₄⁺] = [OH⁻] = X</em>
<em>And as </em>[NH₃] = 0.619M
1.8x10⁻⁵ = [X] [X] / [0.619M]
1.11x10⁻⁵ = X²
3.34x10⁻³ = X = [NH₄⁺] = [OH⁻]
<h3>[OH⁻] = 3.34x10⁻³M</h3><h3 />
% ionization:
[NH₄⁺] / [NH₃] * 100 = 3.34x10⁻³M / 0.619M * 100 = 0.54%
pH:
As pOH = -log [OH-]
pOH = 2.48
pH = 14 - pOH
<h3>pH = 11.52</h3>
Answer 10
Explication nose
