Asked by Mike
1. 25mL of 0.1M HCl is titrated with 0.1M NaOH. What is the pH when 30mL of NaOH have been added?
2. 25mL of 0.1M HCl is titrated with 0.1M NaOH. What is the pH after 15mL of NaOH have been added?
I don't know how to make calculation I assumed that for number 1 when you have added 25 mL of NaOH the pH would be neutral (7) but after you add the rest of the 5mL it would be greater than 7. However, we're suppose to give a specific pH value.
#2 I guessed that the pH would be very low. But how do make the calculations?
2. 25mL of 0.1M HCl is titrated with 0.1M NaOH. What is the pH after 15mL of NaOH have been added?
I don't know how to make calculation I assumed that for number 1 when you have added 25 mL of NaOH the pH would be neutral (7) but after you add the rest of the 5mL it would be greater than 7. However, we're suppose to give a specific pH value.
#2 I guessed that the pH would be very low. But how do make the calculations?
Answers
Answered by
DrBob222
You're thinking is on the right track for #1. #2 is the same kind of problem.
For both, calculate moles of HCl initially, (moles = M x L), then calculate moles NaOH added. In the first case, the excess hydroxide will be in moles, divide that by the total volume and that give you molarity of OH^-. Calculate pOH and pH from that.
For #2 it's just the reverse. moles HCl initially - moles OH added leaves an excess of HCl unreacted, moles HCl remaining divided by total volume = molarity and from there you get pH. Post back if you get stuck.
For both, calculate moles of HCl initially, (moles = M x L), then calculate moles NaOH added. In the first case, the excess hydroxide will be in moles, divide that by the total volume and that give you molarity of OH^-. Calculate pOH and pH from that.
For #2 it's just the reverse. moles HCl initially - moles OH added leaves an excess of HCl unreacted, moles HCl remaining divided by total volume = molarity and from there you get pH. Post back if you get stuck.
Answered by
HP
I'm sooo stuck.
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