If 36.0 mL of 7.0e-4 M HClO4 is added to 19.5 mL of 8.2e-4 M LiOH, what is the pH of the solution?
Write the equation.
Calculate mols HCLO4.
Calculate mols LiOH.
Determine how much of which reagent remains after the reaction.
pH = -log(H^+)
post your work if you get stuck.
found the mols of both HClO4 and LiOH
what do you mean by reagent remaining after reaction?
Write the equation. One is an acid, the other a base, they will react to form a salt + H2O. One will be used up and some of the other will remain.
HClO4 + LiOH --> HOH + LiClO4
how do you figure which is used up and which remains?
Look at the problem below (one of your posts) on Ba(OH)2 and HCl. You asked the same question there and I worked the problem in detail. This one is done the same way. Get back to me if you don't understand.
To determine which reagent is used up and which one remains after the reaction, you need to compare the number of moles of each reagent.
First, calculate the moles of HClO4:
Moles of HClO4 = volume of HClO4 (in liters) * molarity of HClO4
Given:
Volume of HClO4 = 36.0 mL = 0.036 L
Molarity of HClO4 = 7.0e-4 M
Moles of HClO4 = 0.036 L * 7.0e-4 M = 2.52e-5 moles
Next, calculate the moles of LiOH:
Moles of LiOH = volume of LiOH (in liters) * molarity of LiOH
Given:
Volume of LiOH = 19.5 mL = 0.0195 L
Molarity of LiOH = 8.2e-4 M
Moles of LiOH = 0.0195 L * 8.2e-4 M = 1.599e-5 moles
Since the stoichiometry of the reaction is 1:1 between HClO4 and LiOH, and the moles of HClO4 (2.52e-5 moles) are greater than the moles of LiOH (1.599e-5 moles), it means that HClO4 will be the limiting reagent and LiOH will be in excess.
To determine the amount of each reagent that remains after the reaction, you need to calculate the moles of the excess reagent that is not used.
Moles of HClO4 remaining = Moles of HClO4 initial - Moles of HClO4 reacted
Moles of HClO4 remaining = 2.52e-5 moles - 1.599e-5 moles = 9.21e-6 moles
Since we now know the moles of HClO4 remaining and the volume of the solution, we can calculate the concentration of H+ ion using pH = -log(H+):
pH = -log[H+]
pH = -log(9.21e-6 / (36.0 mL + 19.5 mL))
Note that you should convert the volumes to liters before calculating the concentration.
If you get stuck or want to double-check your work, feel free to post your calculations.