Asked by Rey
1.76g of sodium hydroxide was added to 50cm^3 of 0.025mol.l^-1 nitric acid. Calculate the pH of the resulting solution.
Answers
Answered by
bobpursley
NaOH+HNO3>>NaNO3 + water
moles acid= .050dm^3*.025mol/dm^3
= 1.25mMole (check that).
Moles NaOH=1.76/40= 44mMole (check that).
moles of NaOH in excess:44-1.25=you do it.
This represents the OH concentration...
OH=(molesOHabove/.05)
H= 1E-14/OHabove
pH= -log(H)
moles acid= .050dm^3*.025mol/dm^3
= 1.25mMole (check that).
Moles NaOH=1.76/40= 44mMole (check that).
moles of NaOH in excess:44-1.25=you do it.
This represents the OH concentration...
OH=(molesOHabove/.05)
H= 1E-14/OHabove
pH= -log(H)
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.