Consider the titration of 60.0 mL of 0.150 M HNO3 with 0.450 M Na...

Question

Consider the titration of 60.0 mL of 0.150 M HNO3 with 0.450 M NaOH.
(a) How many millimoles of HNO3 are present at the start of the titration?
(b) How many milliliters of NaOH are required to reach the equivalence point?
(c) What is the pH at the equivalence point? 
(d) Sketch the general shape of the pH titration curve.

Accepted Answer

Hello everyone today we have the following problem and an experiment 50 mL of 500.200 molar hydrochloric acid is tight treated with 0.400 moller of potassium hydroxide a says to calculate the initial moles of hydrochloric acid that is present in the solution. So polarity is measured in moles per liter. So we can do to solve for the moles. We can use the formula that we have the malaria T. Which is equal to r 0.200 moles per one liter. We can then multiply by our volume which is 50 middle leaders And then we have to convert those millions to leaders and we do that by using the conversion factor that one mil. A leader is equal to 10 to the negative third leaders When our units canceled out, we're left with 0.01 moles of our hydrochloric acid. And this is the initial number of moles that is present. Now we move on to be B says to calculate the volume of 0. molar sodium hydroxide required to reach the equivalence point for that. We're gonna use the formula the polarity times the volume for our acid is equal to the polarity of our base times the volume of our base Plugging in our values. We have the malaria for asset which was 0.200 moller The volume of the acid was 50 ml. The polarity of our base was 0.400 moller and we don't have the volume for our base. When we saw for the volume of our base We get 25 miller leaders moving on to see it says, determine the ph of the solution at the equivalence point. Well, first, let's take a look at what we have. We have hydrochloric acid, which is a strong acid, and we have potassium hydroxide, which is a very strong base. And when you have a strong acid, strong base titrate in the ph at the equivalence point Is equal to seven. So the ph at the equivalence point is equal to seven. Last but not least, we have d identify the ph titrate curve for this experiment. So, a titillation curve is going to have the usual Y axis and the X axis, we're gonna label zero at the corner there, we're gonna have our volume at the bottom. And middle leaders, we're gonna have our ph on the Y axis, we're gonna have seven and 14 as our ph labels and we're gonna have 25 is our midway point. And 50 mL. The titillation curve is going to show a curve. And then when we get to 25 mm, we're going to draw a straight arrow up or straight line up and then plateau when we get to 14. And so why is this? Well that is because For B, we already saw that at 25 ml, that's when we reached the equivalent point. And so that's why we have this resulting graph here. And ultimately, we have solved this problem. I hope that this helped. And until next time

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