If a 0.200M solution of an acid, HX, has a pH=1.00, The Ka=
a. 2 x 10^-1
b. 2 x 10^-2
c. 1
d. 1 x 10^-1
e. 1 x 10^-3
This is what I did:
HX <===> H + X
10^-1=[H]
0.1=[H]
[H]=[X]
Ka = (H)(X)/HX
Ka = (0.1)(0.1)/0.200
Ka = 0.05
HX should = .2-.1, right?
so Ka=.01/.1=.1
check my thinking
Seems correct to me
To calculate the Ka value of the acid, we need to use the given information that the 0.200M solution of the acid has a pH of 1.00. First, convert the pH to the concentration of hydronium ions [H+].
The pH is defined as the negative logarithm (base 10) of the concentration of hydronium ions. Therefore, we have:
pH = -log[H+]
Given that the pH is 1.00, we can rewrite the equation as:
1.00 = -log[H+]
To solve for [H+], we need to evaluate the inverse logarithm, which is 10 raised to the power of (-1.00):
[H+] = 10^(-1.00)
[H+] = 0.1
Since the solution is dilute, we can assume that the concentration of HX is approximately equal to the concentration of hydronium ions [H+].
Using the equation for Ka, where Ka is the acid dissociation constant:
Ka = ([H+][X-]) / [HX]
We substitute the known values:
Ka = (0.1)(0.1) / 0.200
Ka = 0.05
Therefore, the Ka value of the acid, HX, is 0.05.
Looking at the answer choices given, none of them matches with our calculated value of 0.05 as the Ka value. It is possible that there may be an error in the given answer choices, or further clarification may be required.