00:01
To answer this question, we'll use the henderson -hasselbalch equation that tells us ph, the ph that we desire is 5 .75, will be equal to pca of the acid in the buffer.
00:16
P -k -a is the negative log of the k -a value of acetic acid, 1 .8 times 10 to the negative 5, plus the log of we could do moles base over moles acid or we could do concentration base over concentration acid it's typically easier to use moles so the moles of the base the moles of the sodium acetate is going to be the volume of the buffer solution 0 .620 liters 620 millimeters of the buffer is 0 .620 liters multiplied by the concentration of the sodium acetate, which is 1 liter, is 0 .0245 moles.
01:19
Then when we add our sodium hydroxide, the amount we don't know, we're going to react with the acetic acid reduce the amount of acid we have and increase the amount of acetate we have by the moles of sodium hydroxide we add.
01:39
So it's going to be plus the concentration, well plus the volume, we don't know that'll be x multiplied by the concentration of 6 .01 moles per liter.
01:56
Then we'll divide by the moles of acetic acid, which will be the volume of acetic acid, 0 .620 liters multiplied by its concentration 0 .01 -90 moles per liter minus the volume of sodium hydroxide we add multiplied by its molarity.
02:25
Every mole of sodium hydroxide we add will decrease the moles of acetic acid by the same amount and increase the moles of acetate by the same amount.
02:35
That's because when we add sodium hydroxide expressed as oh minus, it'll react with our acetic acid, making acetate and water in a one -to -one fashion.
02:56
So taking the negative log of this, subtracting it from both sides, we get 1 .005 being equal to the log of 0 .05 being equal to the log of 0 .0 .0 .0 .5...