What is the molarity of OH- in a 5.78×10-3 M C6H5CO2Na solution that hydrolyzes according to the equation.

C6H5CO2-(aq) + H2O(l) = OH-(aq) + C6H5CO2H(aq)

Kb = (Kw/Ka) = (OH^-)(C6H5COOH)/(C6H5COO^-)

Plug in values for Kw, Ka, x for OH and x for C6H5COOH, and 5.78E-3 for C6H5COO^-) and solve for x.

To determine the molarity of OH- in the given solution, we need to first understand the stoichiometry of the reaction and the relationship between the hydrolysis of C6H5CO2Na and the formation of OH-.

From the balanced equation, we can see that 1 mole of C6H5CO2- reacts with 1 mole of water to produce 1 mole of OH-. This means that the moles of OH- produced will be equal to the moles of C6H5CO2Na that undergo hydrolysis.

Given the molarity of the C6H5CO2Na solution is 5.78×10-3 M, we can use this information to calculate the moles of C6H5CO2Na present.

Moles of C6H5CO2Na = Molarity × Volume

Let's say the volume of the solution is V liters. Therefore,

Moles of C6H5CO2Na = (5.78×10-3 M) × V

Since the reaction is 1:1 between C6H5CO2Na and OH-, the moles of OH- will be the same as the moles of C6H5CO2Na.

Now, we convert the moles of OH- to molarity by dividing the moles of OH- by the volume of the solution.

Molarity of OH- = Moles of OH- / Volume = Moles of C6H5CO2Na / Volume
= (5.78×10-3 M × V) / V
= 5.78×10-3 M

Therefore, the molarity of OH- in the given C6H5CO2Na solution is 5.78×10-3 M.