200mL of 0.15 M HCl is added to 300mL of 0.090 M NaOH. What is the pH of the resulting solution?

First, we need to determine which of the two solutions, HCl or NaOH, is the limiting reactant. This is because the reaction between HCl and NaOH is a neutralization reaction, where HCl (a strong acid) and NaOH (a strong base) react to form water and a salt (NaCl). The limiting reactant is the one that will be completely used up in the reaction.

The number of moles of each reactant can be calculated using the formula:

Moles = Molarity x Volume (in liters)

For HCl: Moles of HCl = 0.15 M x 0.2 L = 0.03 moles

For NaOH: Moles of NaOH = 0.09 M x 0.3 L = 0.027 moles

Since NaOH has fewer moles, it is the limiting reactant. All of the NaOH will react with the HCl, leaving some HCl unreacted.

The moles of HCl left over after the reaction can be calculated by subtracting the moles of NaOH from the moles of HCl:

Moles of HCl left = 0.03 moles - 0.027 moles = 0.003 moles

The molarity of the HCl left in the solution can be calculated by dividing the moles of HCl left by the total volume of the solution (in liters). The total volume of the solution is the sum of the volumes of the HCl and NaOH solutions:

Total volume = 0.2 L + 0.3 L = 0.5 L

Molarity of HCl = 0.003 moles / 0.5 L = 0.006 M

Since HCl is a strong acid, it will completely dissociate in water, so the concentration of H+ ions in the solution will be equal to the molarity of the HCl. The pH of the solution can be calculated using the formula:

pH = -log[H+]

pH = -log(0.006) = 2.22

So, the pH of the resulting solution is 2.22.