Calculate the pH of a solution when adding 30.00 mL of 0.1000 M NaOH on 50.00 ml of 0.0500 M HCl with at 25°C.

*
1 point
11.80
11.2
8.0
4.0

Bot GPT 3.5 Bot GPT 3.5 answered
1 month ago

The first step is to determine the moles of NaOH and HCl used in the reaction.

moles NaOH = 0.030 L * 0.1000 mol/L = 0.003 moles NaOH
moles HCl = 0.050 L * 0.0500 mol/L = 0.0025 moles HCl

Since NaOH and HCl react in a 1:1 stoichiometric ratio, the NaOH will completely react with the HCl, leaving no excess NaOH. This means that the resulting solution will only contain the products of the reaction, which are water (H2O) and sodium chloride (NaCl).

The total volume of the solution after the reaction is 30.00 mL + 50.00 mL = 80.00 mL = 0.0800 L

Since both NaOH and HCl are strong electrolytes, they will dissociate completely in solution. The resulting solution will contain water and salt (NaCl), which will not affect the pH.

The pH of the resulting solution can be calculated by determining the concentration of OH- ions left in solution after the reaction, which will come from the excess NaOH that did not react with the HCl.

After the reaction, there are 0.003 moles - 0.0025 moles = 0.0005 moles of NaOH left in solution.

The concentration of NaOH in the resulting solution is
0.0005 moles / 0.0800 L = 0.00625 mol/L

Since NaOH is a strong base, it will dissociate completely in solution to form equal concentrations of Na+ and OH- ions. Therefore, the concentration of OH- ions in the solution is also 0.00625 mol/L.

Taking the negative logarithm of the OH- concentration gives the pOH of the solution
pOH = -log(0.00625) = 2.20

Finally, the pH of the solution can be calculated as
pH = 14 - pOH = 14 - 2.20 = 11.80

Therefore, the pH of the resulting solution after mixing 30.00 mL of 0.1000 M NaOH with 50.00 mL of 0.0500 M HCl at 25°C is 11.80.

So, the correct answer is:
11.80