Determine the ph of a 0.00598 m hclo4 solution. 1.777 6.434 7.566 2.223 3.558
2 answers:
pH is the degree of acidity of a solution that depends on the concentration of H⁺ ions. The greater the value the more acidic the solution and the smaller the pH.
pH = - log [H⁺]
So that the two quantities between pH and [H⁺] are inversely proportional because they are associated with negative values.
A solution whose value is different by n has a difference in the concentration of H ions of 10ⁿ.
0.00598 M HClO₄ solution
HClO₄ is a strong ionized acid:
HClO₄ ---> H⁺ + CLO₄⁻
produces 1 H⁺ ion so it has acid valence 1
So we use a strong acid formula to find the pH
[H⁺] = a. M (a = acid valence, M = acid concentration)
[H⁺] = 1. 0.00598
[H⁺] = 5.98 .10⁻³
pH = - log [H⁺]
pH = - log 5.98 .10⁻³
pH = 3- log 5.98
pH = 2,223
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the pH of a solution
Calculate the pH
the pH of a 2.0 M solution of HClO₄
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Answer:
<u>D. It will decrease by a factor of 4</u>
Explanation:
According to the question , the equation follows :
Rate law : This states the rate of reaction is directly proportional to concentration of reactants with each reactant raised to some power which may or may not be equal to the stoichiometeric coefficient.
.................(1)
STEP": First, find out the power "a" and "b"
a+b = 3 (because it is given that the reaction follow 3rd order-kinetics)
According to question, <u><em>doubling the concentration of the first reactant causes the rate to increase by a factor of 2 means,</em></u>
r' = 2r if [A'] = 2[A]
Here [B] is uneffected means [B']=[B]
hence new rate =
Put the value of [A'] , [B'] and r' in the above equation:
...........(2)
Divide equation (1) by (2) we , get
Here A and A cancel each other
B and B cancel each other
We get,
1^b = 1 ( power of 1 = 1)
This is possible only when a = 1
We know that : a + b = 3
1 + b = 3
b =3 -1 = 2
b = 2
Hence the rate law becomes :
<u>.............(3)</u>
Look in the question now, it is asked to calculate the concentration of [B],if cut in half
Hence
[B']=1/2[B]
Insert the value of [B'] in equation (3)
............(a)
But
..............(b)
Compare equation (a) and (b) , we get
new rate r' =
<u>r' = 1/4 r</u>