For the reaction show below determine the rate law. H2SO3(aq) + 6I-(aq) + 4H+(aq) → S(s) + 2I3- + 3H2O (l) In the presence of a
little starch, this solution turns from colorless to blue when I3- is formed. The following data were obtained: [H2SO3] [H+] [I-] Initial Rate (mol/L s) 1.0 x 10-2 2.0 x 10-1 2.0 x 10-1 1.66 x 10-7 3.0 x 10-2 2.0 x 10-1 2.0 x 10-1 4.98 x 10-7 1.0 x 10-2 1.0 x 10-1 2.0 x 10-1 8.30 x 10-8 1.0 x 10-2 2.0 x 10-1 4.0 x 10-1 13.2 x 10-7 Calculate the rate law for this reaction: rate = k [H2SO3] [H+] [I-
1 answer:
You might be interested in
Hi , I’m really sorry because I just have the first question... so here it is :
So it is 0.52 mol .
Have a nice day .
Answer:
- <em>The solution that has the highest concentration of hydroxide ions is </em><u>d. pH = 12.59.</u>
Explanation:
You can solve this question using just some chemical facts:
- pH is a measure of acidity or alkalinity: the higher the pH the lower the acidity and the higher the alkalinity.
- The higher the concentration of hydroxide ions the lower the acidity or the higher the alkalinity of the solution, this is the higher the pH.
Hence, since you are asked to state the solution with the highest concentration of hydroxide ions, you just pick the highest pH. This is the option d, pH = 12.59.
These mathematical relations are used to find the exact concentrations of hydroxide ions:
- pH + pOH = 14 ⇒ pOH = 14 - pH
- pOH = - log [OH⁻] ⇒
Then, you can follow these calculations:
Solution pH pOH [OH⁻]
a. 3.21 14 - 3.21 = 10.79 antilogarithm of 10.79 = 1.6 × 10⁻¹¹
b. 7.00 14 - 7.00 = 7.00 antilogarithm of 7.00 = 10⁻⁷
c. 7.93 14 - 7.93 = 6.07 antilogarithm of 6.07 = 8.5 × 10⁻⁷
d. 12.59 14 - 12.59 = 1.41 antilogarithm of 1.41 = 0.039
e. 9.82 14 - 9.82 = 4.18 antilogarithm of 4.18 = 6.6 × 10⁻⁵
From which you see that the highest concentration of hydroxide ions is for pH = 12.59.