Question

A chemist mixes 50.0mL of a 1.0M NaOH solution with 50.0mL of a 1.0M Ba(OH)2 solution. Assuming the two solutions are additive, what is the pH of the resulting solution

Answer 1

`pH=14.2`

Step-by-step explanation:

Hello there!

In this case, according to the information in this problem, and considering these two bases are strong, it is necessary for us to calculate the total moles of OH ions as shown below:

`n_(OH^-)^(from\ NaOH)=0.050L*1.0mol/L=0.050mol\\\\n_(OH^-)^(from\ Ba(OH)_2)=0.050L*1.0mol/L*2=0.10mol\\\\n_(OH^-)^(tot)=0.15mol`

Now, the as the solutions are additive, the total volume is then 0.100 L and the concentration:

`[OH^-]=(0.15mol)/(0.100L)=1.5`

And therefore, the pH is:

`pH=14+log(1.5)\\\\pH=14.2`

Regards!