# 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 -> C6H6COO- + OH-}\frac{0.0375 ~\mathrm{mol}}{.100~ \mathrm{L}} = 0.375~M\ce{pOH} = 0.43\ce{pH} = 13.57$$Is this right? Thank you. • Benzoate is a weak base, so you can't assume it reacts completely to form \ce{OH^{-}}. What's the data you need in this case? Also, your reaction is written a bit incorrectly. – Nicolau Saker Neto Apr 27 '15 at 1:08 • Would you need Kb? – user15831 Apr 27 '15 at 1:13 ## 1 Answer Welcome to Chemistry SE! To do this specific problem you need to use the K_{b}:$$\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.