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What will be the ph of 1m CH3SO3H? ka of CH3SO3H = -1.9
Accepted Answer
The pH of a 1M solution of CH3SO3H can be calculated using the following equation:
pH = -log[H+]
where [H+] is the molar concentration of hydrogen ions in the solution.
The Ka of CH3SO3H is -1.9, which means that it is a weak acid. The equilibrium constant expression for the dissociation of CH3SO3H is:
CH3SO3H + H2O CH3SO3- + H3O+
Ka = [CH3SO3-][H3O+]/[CH3SO3H]
Assuming that the initial concentration of CH3SO3H is 1M, and that the concentration of CH3SO3- and H3O+ is negligible, we can simplify the equilibrium constant expression to:
Ka = [H3O+]^2/[CH3SO3H]
Solving for [H3O+], we get:
[H3O+] = sqrt(Ka * [CH3SO3H])
Plugging in the values for Ka and [CH3SO3H], we get:
[H3O+] = sqrt(-1.9 * 1M) = 0.43M
Now we can calculate the pH:
pH = -log(0.43M) = 0.37
Therefore, the pH of a 1M solution of CH3SO3H is 0.37.
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