A solution is made by adding 0.300 g Ca1OH221s2, 50.0 mL of 1.40 ...

Question

A solution is made by adding 0.300 g Ca1OH221s2, 50.0 mL of 1.40 M HNO3, and enough water to make a final volume of 75.0 mL. Assuming that all of the solid dissolves, what is the pH of the final solution?

Accepted Answer

Hello everyone today. We are being given the following problem. 1.82 g of magnesium hydroxide is dissolved in 63.9 mL of 0.55 molar of hypo chloride acid to make a solution with the final volume of 65 mL, calculate the ph of the final solution if all the solids are dissolved. So the first thing you want to make note of is what we are working with. So we have our magnesium hydroxide here, which is a strong base, it's a strong base and then we have our chloride here, which is a strong acid. And so the next thing we need to do is we want to find, we need to find the moles of each are acid in our base. So far moles of our base, we're going to take our Mass 1. g of it. We're going to multiply by its molar mass that we can find the moles of it. And the molar mass of this is 58.33 g per the periodic table. And that gives us 0.0312 moles. And for acid we're going to follow a very similar procedure. But instead of taking the grams, we automatically have the polarity. So we're gonna take our 0.55 Mosul leaders which does equal polarity multiply that by our nine millimeters from the question stem and then we're going to convert that into leaders. We're going do one leader 0 to 1000 militia leaders. And so our units are going to cancel that. We're gonna be left with 0.0351 moles. Notice how our moles of our acid is greater than the moles of our base. And so what does this indicate? This indicates that the solution is acidic. So our solution here is acidic. So to find out how many much excess that we have, we're gonna take our Moldova acid .0351 moles and subtract bowls of base which is .0312 moles. To be left with 0.0039 malls of excess protons or hydro ni um concentration. So to find the hydrogen concentration, we're simply going to take the number of moles that we have which is 0.39 moles that we just calculated And divide that by the number of leaders that we have. So we have 65 ml which we must convert to leaders And use the conversion factor that one leader is equal to 1000 ml, arguments can cancel out and we can be left with Our concentration of .06 polarity and last but not least to find that ph we must do the negative log Of our hydrogen concentration which equals the negative log. But we just calculated of 0.06 giving us a final ph of 1.2-1 8 as our final answer. Overall. I hope this helped. And until next time

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