-1
$\begingroup$

Calculate the $\ce{pH}$ of a $100$ $\ce{mL}$ solution containing $0$$.0375$ moles of the weak base, Sodium Benzoate, $\ce{C6H5COONa}$.

$$\ce{C6H5COO + H2O -> C6H6COO- + OH-}$$

$$\frac{0.0375 ~\mathrm{mol}}{.100~ \mathrm{L}} = 0.375~M$$ $$\ce{pOH} = 0.43$$ $$\ce{pH} = 13.57$$

Is this right? Thank you.

$\endgroup$
2
  • $\begingroup$ Benzoate is a weak base, so you can't assume it reacts completely to form $\ce{OH^{-}}$. What's the data you need in this case? Also, your reaction is written a bit incorrectly. $\endgroup$ Commented Apr 27, 2015 at 1:08
  • $\begingroup$ Would you need Kb? $\endgroup$
    – user15831
    Commented Apr 27, 2015 at 1:13

1 Answer 1

3
$\begingroup$

Welcome to Chemistry SE!

To do this specific problem you need to use the $K_{b}$:

$$\ce{\frac{x^{2}}{.375-x}}=K_{b}$$

Finding the $K_{b}$ and solving for x will give you the concentration of $\ce{OH-}$. Remember to take the $-log$ of that and solve for the $pH$.

Also next time please show some work and take your time to format these questions so people can try to reply back to you.

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.