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Ch.4 - Reactions in Aqueous Solution

Chapter 4, Problem 82a

(a) How many milliliters of 0.120 M HCl are needed to completely neutralize 50.0 mL of 0.101 M Ba(OH)2 solution?

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Hi everyone is here as the volume of a sighting hydroxide solution that is needed to completely new tries a solution that has nitric acid and phosphorous acid recall that polarity, is he going to malls of the salute. About about leaders of the solution? Since we're given male leaders we need to convert to leaders Left 250 male leaders And in 1000 Male leaders have one leader And this will give us .35 leaders. And now we need to find the moles of nitric acid and phosphorus acid. Then we can add these up to find the total number of moles in the solution. So I have .35 leaders And we have 0.12 mauler of nitric acid which is going to be 0. malls of nitric acid in one leader And it's given 0. malls of nitric acid. And then for Foster's acid it's gonna be try protic Because there's three hydrogen. This is gonna be three Time similarity, which is going to be 0.3 smaller. Would you give us 0.9 smaller. 7.35 leaders In 0.9 Moller a phosphorous acid. It's gonna be 0.9 balls of H three he of four and one leader. And it's to give us 0.3 15 balls phosphorous acid. And now for the total number of moles You have 0.042 malls, 0.315 balls. And this will give us 0.3 57 balls. So now that we have the total number of moles of the solution and the polarity of the study of hydroxide solution, we can find the volume. Using the polarity equation, We're gonna have 0.211 polar Equal to 0.3 57 moles divided by X. We're gonna do 0.211 baller times X. And this equal to 0.357 malls. So if you solve for X And divide both sides by 0. smaller, we're gonna get X. It goes 1.69 leaders. Thanks for watching my video and I hope it was helpful.