Can Someone Help Me With These 2 Chem. Problem About pH?

Question

5.54 mL of 0.0557 M HCl is added to 3.37 mL of 0.0377 M NaOH. The pH of the resultant solution is______

5.42 mL of an HCl solution whose pH is 1.84 is added to 3.21 mL of an NaOH solution whose pH was 9.39. The pH of the resultant solution is _________

Answer 1

First figure out how many moles of H+ come from the HCl.
Since the HCl-H+ ratio is 1:1 use the volume and molarity of the HCl and the answer will be moles of H+.

Use Molarity = moles/liters to get that answer. Don't forget the change mL to liters.

Then figure out how many moles of OH- come from the NaOH using the same formula. The NaOH - OH- ratio is also 1:1.

One of the two numbers, H+ or OH-, is bigger than the other; subtract to find out which one is in excess. If it is the H+, get the pH by taking the -log. If it is the OH-, get the pOH by taking the
-log and then subtract from 14 to get the pH
(since pH + pOH = 14).

Second problem: take the antilog of -1.84 to get the H+.
For the OH- subtract the pH from 14 to get the pOH and then take the antilog of that number to get the OH- (remember to change the number to a negative before you take the antilog).

You now have your H+ and OH-, proceed as in the first problem

Answer 2

you're able to desire to ascertain what share moles you have. what's the reaction happening? (NH4+) + (OH-) -> (NH3) for the reason which you start up with a base (ammonia) and are including some acid. Your pH could desire to drop from the unique pH. So, initially you had 2.0L of .200M NH3. what share moles is that? .200M = .2 hundred moles / one million L So 2.0L * (.2 hundred moles / one million L) = .4 hundred moles NH3 Your pH tells you the way many moles of OH- you have. additionally, undergo in concepts that this could be a vulnerable acid/base pair. there is one extra step, yet i don't elect to grant all of it to you. desire it grew to become into sufficient to get you kick started.