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Question a.
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Calculate the ph during the titration of 50 ml of 0 .1 molar hydrochloric acid with 0 .1 molar of sodium hydroxide solution after adding 50 .1 ml of base.
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So first of all, we know the reaction is hcl plus naoh and then they neutralize each other to produce sodium chloride and water.
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So they have a 1 to 1 molar ratio.
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So now we have a sodium, sorry, hydrochloric acid.
00:41
It has 0 .1 molar and the volume is 50 ml.
00:48
So the total mole of hydrochloric acid equals the volume times the concentration.
01:01
So the volume is 50 ml.
01:06
Concentration is 0 .1 molar.
01:09
The molar can also be converted into mole per liter.
01:15
Now you have a ml here, you have ml here.
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So first of all, 1 ml is 0 .001 liter.
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So 50 is timing 50 on both sides, you get 0 .05 liter.
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So this replaces, the 50 is 0 .05 liter times 0 .1 mole per liter, 0 .005 mole.
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So this is what we have for the hydrochloric acid.
02:00
The same way we can calculate the mole of sodium hydroxide, same way.
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So the mole equals the volume.
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Now this is 50 .1 ml and the concentration is 0 .1 mole per liter, which is molar.
02:27
So then we know that 50 .1 molar, ml converted to liter is 0 .0501 liter.
02:39
So this is the conversion and times 0 .1 mole per liter, you get 0 .00501 mole.
02:57
So if you take a compare these two, you can see that you have additional 0 .00001 mole of sodium chloride that is not being used because 0 .00501 mole minus the amount of the one that used to neutralize hydrochloric acid, and you will have additional 0 .00001 mole.
03:46
Now this is the additional oh that is left in the solution.
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So now we have a total volume of 100 .1 ml.
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So this is the volume of hydrochloric acid and this is the volume of sodium hydroxide.
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So you get 100 .1 ml...