Question

PH CHEM, PLEASE HELP QUICK! NO LINKS/VIRUSES PLEASE!

Answer 1

Answer:

The pH of the solution is 11.48.

Explanation:

The reaction between NaOH and HCl is:

NaOH  +  HCl  →  H₂O  +  NaCl

From the reaction of 3.60x10⁻³ moles of NaOH and 5.95x10⁻⁴ moles of HCl we have that all the HCl will react and some of NaOH will be leftover:

$n_{NaOH}} = n_{i_{NaOH}} - n_{HCl} = 3.60 \cdot 10^{-3} moles - 5.95 \cdot 10^{-4} moles = 3.01 \cdot 10^{-3} moles$

Now, we need to find the concentration of the OH⁻ ions.

$[OH^{-}] = \frac{n_{NaOH}}{V}$

Where V is the volume of the solution = 1.00 L                

$[OH^{-}] = \frac{n_{NaOH}}{V} = \frac{3.01 \cdot 10^{-3} moles}{1.00 L} = 3.01 \cdot 10^{-3} mol/L$

Finally, we can calculate the pH of the solution as follows:

$pOH = -log([OH^{-}]) = -log(3.01 \cdot 10^{-3}) = 2.52$

$pH + pOH = 14$

$pH = 14 - pOH = 14 - 2.52 = 11.48$

Therefore, the pH of the solution is 11.48.

I hope it helps you!