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What will be the ph of 0.9m CH3SO3H? ka of CH3SO3H = -1.9
Accepted Answer
The pH of 0.9M CH3SO3H can be calculated using the Henderson-Hasselbalch equation:
pH = pKa + log([A-]/[HA])
where pKa is the negative logarithm of the acid dissociation constant, [A-] is the concentration of the conjugate base, and [HA] is the concentration of the acid.
For CH3SO3H, the pKa is -1.9, and the concentration of the acid is 0.9M. The concentration of the conjugate base can be calculated using the dissociation constant:
Ka = [H+][A-]/[HA]
Substituting the values we know into this equation, we get:
-1.9 = [H+][A-]/0.9
Solving for [A-]/0.9, we get:
[A-]/0.9 = 10^-1.9
[A-]/0.9 = 0.0126
[A-] = 0.01134M
Now we can plug this value into the Henderson-Hasselbalch equation:
pH = -1.9 + log(0.01134/0.9)
pH = -1.9 + (-1.848)
pH = -3.748
Therefore, the pH of 0.9M CH3SO3H is approximately -3.748.
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