00:01
We recognize that the reaction is going to be one mole of hcl reacts with one mole of sodium hydroxide to produce one mole of sodium chloride and water.
00:17
So if we have 40 milliliters of hcl at a concentration of 0 .100 molar and the sodium hydroxide concentration is also 0 .100 molar, then we're going to need 40 milliliters naoh to reach the equivalence point where there's no more hcl and no more sodium hydroxide.
00:47
They both reacted with each other producing a neutral salt and water where the ph is 7.
00:54
So if we haven't quite added 40 milliliters, say we've only added 35 milliliters, then what's in excess is hcl and we can calculate the ph by first determining the excess moles of hcl.
01:10
The excess moles of hcl will come from the excess milliliters.
01:14
We started with 40.
01:18
We've added 35 milliliters naoh.
01:23
So that means 5 milliliters hcl did not react.
01:27
So the hcl is 0 .005 liters.
01:34
We multiply that by the concentration of hcl, 0 .100 moles per liter, and this gives us the moles of hcl in excess.
01:45
We then divide that by the new total volume.
01:48
The 40 milliliters we started with plus the 35 milliliters of naoh gives us 75 milliliters or 0 .0750 liters.
02:00
And we get an excess hcl concentration of 6 .66 repeating times 10 to the negative 3 molar...