A 40.0 mL sample of a mixture of HCl and H3PO4 was titrated with 0.100 M NaOH. The first equivalence point was reached after 88.0 mL of base, and the second equiva-lence point was reached after 126.4 mL of base. (b) What are the initial concentrations of HCl and H3PO4 in the mixture?

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Welcome back everyone. A 0.1 50 molar sodium hydroxide solution was used to titrate a 50 millimeter sample containing hydrochloric acid and arsenic acid. 78.9 mL of sodium hydroxide was required to reach the first equivalence 0.138 point two mL of sodium hydroxide was required to reach the second equivalence point determined the initial concentrations of hydrochloric acid and arsenic acid in the mixture focusing on our first equivalence point. First, in this tight relation, we want to recall that all moles of hydro ni um both from the first ionization of arsenic acid H three A S 04. And from our hydro ni um from hydroponic acid, both moles of hydro ni um are going to be consumed. We want to recall that at the first equivalence point, our moles of acid should equal our moles of base. So in this case, we can say that therefore, our moles of protons from our strong acid Hydrofluoric acid plus our moles of protons from the first equivalence point of arsenic acid should equal our moles of hydroxide from our base sodium hydroxide. So we need to determine our moles of hydroxide from sodium hydroxide. Using the given information from the prompt. At the first equivalence point, we have 78.9 ml of sodium hydroxide. We're going to multiply by the conversion factor to go from mill leaders in the denominator to leaders in the numerator. Since based on the concentration were given for sodium hydroxide. Behalf units of polarity recall that polarity is interpreted as moles per leader. So we recall our conversion factor where our prefix milli tells us that for one leader, we have 10 to the third power milliliters canceling out male leaders were left with leaders. And now we're going to multiply by incorporating are given molar concentration from the prompt as 100.150 moller, which actually we're interpreting as 0.150 moles of sodium hydroxide per leader of solution. Now canceling out leaders with leaders in the denominator. We're left with moles of sodium hydroxide, but we need to multiply by a conversion factor to go from moles of sodium hydroxide in the denominator, two moles of hydroxide in the new waiter oh H minus. So within our formula, we can see that we have a 1 to 1 molar ratio between hydroxide and sodium hydroxide. So canceling out moles of sodium hydroxide. We're left with molds of hydroxide as our final unit. And this is going to simplify to our molds of hydroxide at the first equivalence point equal to a value of 0. moles of hydroxide. Now focusing on what occurs at our second equivalence point. We would recall that almost of hydro ni um from the second ionization of arsenic acid are consumed. And so we can next say that our moles of protons from arsenic acid from its second ionization would equal the moles of hydroxide from sodium hydroxide. And so to find our moles of hydroxide at the second equivalence point from sodium hydroxide, we have our volume. Now being 138.2 mL as given from the prompt of sodium hydroxide. Again multiplied to go from milliliters to liters. In the numerator, we have one leader equivalent to 10 to the third power milliliters in our denominator canceling out milliliters. We will now multiply by our given molar concentration For sodium hydroxide which we interpret as 0.150 moles of sodium hydroxide given from the prompt per liter of solution. So canceling out leaders, we are left with moles of sodium hydroxide. In which again, we go from moles of sodium hydroxide in the denominator, two moles of hydroxide in the numerator using our 1 to 1 molar ratio and canceling out moles of sodium hydroxide. Moles of hydroxide is our final unit. And so we'll find that at the second equivalence point are moles of hydroxide from sodium hydroxide is now 0.2073 moles of hydroxide. So we can conclude that for our weak acid arsenic acid H three A S 04, we can say that the moles of protons required to reach the first equivalent point Is equal to two times the moles of protons required to reach The 2nd equivalence point. So with this understanding, we can now say that our moles of protons from the first equivalent point of arsenic acid, Which we determined above to be 0. moles of hydroxide Are now going to be divided by two. And this is going to result in a value of 0.0103, moles. So now with these calculations figured out, we can calculate next our concentration of protons from hydroponic acid which we can find by taking our moles of hydroxide from our first equivalence point subtracted from our moles of hydrogen or sorry, our moles of protons from the first equivalence point of arsenic acid. And so plugging in our answers from above our moles of hydroxide from the first equivalence point, we determined above as 0.11835 moles of hydroxide Subtracted from our moles of protons from the first equivalence point of arsenic acid, which we just determined above as 0.010365 moles. And so now from this difference, we find our concentration of protons from hydroponic acid equal to a value of 0.1 47 moles. And Now, we can move on to our last step which determine our initial concentration of first our strong acid hydrochloric acid in which we take the calculation of protons being 0.1 moles. And we want to Get a concentration of our final answer. So we're going to divide this by the volume given for our sample which according to the prompts, our sample is a volume of 50 ml. And recall that as we stated, molar concentration or polarity is in terms of moles per liter. So we multiply by the conversion factor where we recall that We have 10 to the third power ml equivalent to one leader. So this will allow us to cancel out our units of male leaders leaving us with moles per liter for our concentration. And we'll find our concentration of hydrochloric acid initially is equal to 0.02 94 Moller. Now, for our initial concentration of arsenic acid, we take our calculation for the moles of protons from the first equivalence point at 0. moles. And we're going to divide this by the molds of our sample or volume of our sample being 15 mL as given from the prompt and multiplied by the same conversion factor where 10 to the third power middle leaders in the numerator is equivalent to one leader in the denominator. Canceling out male leaders were left with moles per liter as our final units and we'll get our concentration of arsenic acid initially of 0.2073 molar. So our two final answers highlighted in yellow will correspond to choice being the multiple choice as our final answer for the initial concentrations of HBR and arsenic acid. In art illustration. I hope this made sense and let us know if you have any questions.

A 40.0 mL sample of a mixture of HCl and H3PO4 was titrated with 0.100 M NaOH. The first equivalence point was reached after 88.0 mL of base, and the second equiva-lence point was reached after 126.4 mL of base. (b) What are the initial concentrations of HCl and H3PO4 in the mixture?

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Key Concepts

Titration

Equivalence Point

Stoichiometry