00:01
To answer this question, we recognize that the stoichiometry between nitrous acid and sodium hydroxide is one -to -one, where one mole nitrous acid reacts with one mole sodium hydroxide to produce one mole sodium nitrite and one mole of water.
00:18
So as long as we don't add more moles of sodium hydroxide than we have nitrous acid, we'll still have some nitrous acid and we would have created some conjugate base nitrite.
00:31
So we have a buffer solution.
00:33
Ph can then be calculated for a buffer solution using the henderson -hasselbalch equation, where ph equals pka, which is the negative log of the ka value, 4 .5 times 10 to the negative 4, plus the log of the moles of the nitrite that is formed.
00:53
Every mole of sodium hydroxide we add produces a mole of nitrite.
00:58
We added 25 milliliters, that's 0 .025 liters at a concentration of 0 .01 moles per liter.
01:08
So those moles of sodium hydroxide added equal the moles of sodium nitrite formed...