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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.

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  • $\begingroup$ 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. $\endgroup$ – Nicolau Saker Neto Apr 27 '15 at 1:08
  • $\begingroup$ Would you need Kb? $\endgroup$ – user15831 Apr 27 '15 at 1:13
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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.

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