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


  $$\ce{C6H5COO + H2O -&gt; C6H6COO- + OH-}$$


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

Is this right? Thank you.
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.