A 20.0-mL sample of 0.125 M HNO3 is titrated with 0.150 M NaOH. Calculate the pH for at least five different points throughout the titration curve and sketch the curve. Indicate the volume at the equivalence point on your graph.

Video transcript

Welcome back. Everyone below is an incomplete curve for the titration of 25 mL of 250.100 molar chloric acid with 0.1 25 molar potassium hydroxide. The points corresponding to the following volumes of 250.1 25 molar potassium hydroxide are missing zero mL and 10 millis plot the points sketch the curve and label the equivalence point on the graph. Currently, in our given P H curve on our Y axis, we have our P H values ranging up to 14. And then on our X axis, we have our volumes of 140.1 25 molar potassium hydroxide, which is our tiran that is added between 0 to mL. And we observe that we have three listed P H points. The first being at A P H of around we'll say 12.5. Then we have a P H of close to 12 and these two P H points are likely listed at volumes around and over 30 mL of tyrann added. And then our third P H point listed we can say is about at A P H of seven which is neutral and around, we can say maybe 10 to 15 mL of tyrine added at an estimate. Now, let's begin by figuring out which exact volumes are. First two page points that we need to draw in correspond to beginning with 20.1 on our P H curve. We're going to need to understand our reaction. So let's write it out. We have Chloric acid H C L 03, which we should recognize is a strong acid reacting with potassium hydroxide are tirin where potassium hydroxide we recognize as a strong base. So because we have a strong acid and a strong base reacting with one another, we have a neutralization reaction. And on our product side should be a salt and then our second product should be water. So our salt, which forms will be a result of our acid and base will displace one another. So we would be able to write out our ions making up each of our reagents. First with our acid chloric acid, we have our H plus one cat ion for a proton. Then for chlorate C L 03 recall that that carries a one minus charge as a poly atomic ion. Then in our second reactant, potassium hydroxide, our strong base, we have our K plus one potassium ion and then hydroxide, which we recall is a poly atomic ion with a minus one charge. Recall that in a displacement reaction, in this case, we have a double displacement, we would pair up our oppositely charged ions to form our products. So our second product is going to be H2O from the combination of our proton with hydroxide. And our first product, which is our salt is going to be the combination of our potassium ion and our chlorate anion giving us our potassium chloride product. Notice that our charges neutralize one another. So neither of our charges are or neither of our products are charged. And ultimately, this is our entire neutralization and double displacement reaction or double replacement reaction. Now, for 0.1, we are considering the addition of zero mL of our tyrant, potassium hydroxide added. And this means that therefore only our Annaly, which is our substance being tired as our acid, chloric acid is present. So looking at the prompt, we have a volume of chloric acid being 25 mL. So we would begin by noting down that only mL of our chloric acid is present in our solution. And so thus, we are going to recall our formula where we can calculate the ph by taking our negative log of our concentration of protons H plus, which are present in our acid. And therefore, we can plug in our concentration, which is known for our acid and take the negative log of our concentration of 0.100 molar for chloric acid. And in our next line, we would find that our P H is equal to a value of the value one. And this would be our first answer for our P H point at zero mL of tyrant added. So following on our given P H curve on our X axis, we're going to start at our origin and we're going to plot a point at our P H value of one at this location between our P H values of two and zero on the Y axis. Now let's figure out our second P H point point two on our titration curve which considers the addition at mL of our tyrant potassium hydroxide. So now we finally have both of our reagents within solution. Here. In this case, we have 10 mL of potassium hydroxide. So let's figure out how many moles of our base potassium hydroxide that would equal because we understand our polarity for our base, which is given as 0.1 25 molar molar is interpreted in terms of moles per liter. We can solver the moles of our base by taking our molar multiplied by our volume of base used. So this is specifically our initial moles of base before the reaction occurs with our acid. So plugging in our molar concentration given as 0.1 25 molar, which we will interpret as moles per liter. We're then going to multiply by our volume for 250.2 given in our prompt as mL canceling out our units of liters of milliliters. And actually, we can't cancel them out since we're given a volume in millis where you need to convert that to leaders before we can cancel that out. So we need to multiply to go from Millis in the denominator to leaders in the numerator where we recall that our prefix Milly tells us that we have an equivalent of 10 to the negative third of our base unit leader in the numerator. And now canceling out mill leaders along with leaders, we're left with moles as our final unit, which is what we want. And we would find that our initial moles of potassium hydroxide is equal to the value 0.125 moles of potassium hydroxide. Initially. Now let's figure out our initial mold of our strong acid chloric acid where we would again take the molar multiplied by the volume used of our acid. So in this case, from above and sorry, this is an equal sign from above, we determined where we were rather given our molar concentration of our acid as 0.100 molar. So plugging that in and interpreting it as 0.100 moles, we have most per liter. We're then going to multiply by the volume of our acid which above in our prompt is given as 25.0 milit in which we need to convert to liters by multiplying by our conversion factor from before 10 to the negative third liters in the new equivalent to one milli in the denominator again canceling out millis along with leaders, we're left with moles. And we would find that our initial moles of acid is equal to the value 0.25 moles of acid present, Chloric acid H C L 03. Now, we need to determine after our reaction, how many moles of whatever region would be remaining. We can simply look at our initial amounts and determine which reagent is smaller or limiting. However, we can also utilize an ice table or rather an IC F table since we do not have an equilibrium I for initial C for change and F for our final conditions of our reaction. So writing our reaction above, as we stated before, we have chloric acid H C L reacting with potassium hydroxide or tyrant. And a neutralization reaction will be form on the product side. Potassium chloride as our first product and water as our second product recall that in IC F tables or ice charts, even we do not consider our water. We only consider aqueous regents. Now, as far as our initial conditions will plug in our 0.25 moles of our chloric acid, then we can also plug in our 0.1 25 moles of our potassium hydroxide base that we determined. And for our products being our potassium chlorine, we have zero moles of that initially. Since it has not formed yet as a product for the change, we would subtract the limiting reagent moles amount in this case, the smaller amount of moles is associated with potassium hydroxide. So we would have minus 0.1 25 moles from both of our reagents as the change. And we would add this amount for our product. So plus 0.1 25 moles to our product, potassium chloride. And as a result, our final conditions will result in zero moles of our base at the final point of our reaction, we would have as far as our acid, 0.1 25 moles of chloric acid remaining after the reaction and 0.1 moles of potassium chloride remaining after the reaction. So at the final conditions of our reaction, we have our salt and our strong acid remaining where our strong acid will influence P H mainly. So we need to therefore consider our moles that are remaining of our chloric acid H C L 03. So we would take our initial amount 0.1 25 moles. And we want to subtract this from our moles of acid reacted being 0.0 and apologies. Our initial amount of acid is as we wrote above 0.25 moles of chloric acid initially subtracted from our amount of chloric acid reacted 0.1 moles reacted. And so we see that we have 0.1 25 moles that remain as we concluded in our IC F table. So, moles of strong acid that remain and will therefore influence P H. So therefore, we can find our ph of our solution by again, taking the negative log of our concentration of protons in our acid. In this case, we would take the negative log and plug in our concentration of our acid given in our pump as and actually, we need to convert our moles remaining to a concentration. So we would not plug in what we were given. Rather, we need to take our 0.1 25 moles of our strong acid that remains. So let's rewrite that strong acid that remains. And this amount of mold we need to interpret as a concentration. And so we would take our, we would find our polarity of chloric acid in which, as we stated before, molar is in terms of moles of our solute per liter or volume of our solution in liters. So in this case, we would plug in our 0.1 moles of chloric acid H C L 03 that remains. And we want to divide by our volume of our solution in liters by taking our volume of acid used for our reaction, which we stated above is 25 mL. So plugging that in we have 25 mL of Chloric acid H C L 03. This is added to what we have for 30.2 on our P H curve for our volume of base use or tyrant. So plus 10 mL of potassium hydroxide, and then we're going to need to make sure our units are in liters. So we need to multiply our conversion factor in our denominator, so that we have one mL in the denominator equivalent to 10 to the negative third liters in the numerator. And now we can cancel out all of our units of milit leaving us with moles per liter for our units of molar concentration of our acid chloric acid. And in our next line, we can simplify so that we have in our num, we still have 0.1 25 moles of chloric acid. And in our denominator, we can plug in the result of adding and multiplying in our denominator which would be 0. correction. That would be 0.35. And we would have units of liters. And from the result of this quotient, we would then get our molar concentration of chloric acid equal to 0.35714 molar of chloric acid. So this concentration is influencing our P H where, as we stated before, we can take our P H and find the negative log of our concentration of protons H plus in which we would plug in the negative log of our concentration, which we just calculated as 0.35714 molar. And then in our calculators, we would result in a P H equal to about 1.45. And this would be our second answer. So far as our P H when we have the addition of 10 mL of tyrant strong base added. So, plugging this in onto our P H curve for mL of potassium hydroxide, which we'll say is around here. We're here. We have, as we stated 1.45 as our P H. So we would draw our next P H point just a bit higher than our previous point. So we'll say around here. Now, the last step we need to take is labeling our equivalence point. And we want to recall that in a neutralization reaction, our equivalence point is when our moles of acid equal our moles of base because our strong base will have neutralized our strong acid. And so therefore, we would observe that our P H at the equivalence point will be equal to seven. So we can label our ph of seven at this point as our equivalence point for our third answer. And then we just need to connect all of our points to form our P H curve. So connecting our five points, we have our complete P H curve for the reaction between chloric acid and potassium hydroxide. So our entire P H curve along with our labeled equivalence point at a ph of seven represents our final answer to complete this example. I hope that this helped and let us know if you have any questions.