Calculate the pH of solutions prepared by: (d) Mixing 100.0 mL of 2.0 * 10-3 M HCl and 400.0 mL of 1.0 * 10-3 M HClO4. (Assume that volumes are additive.)

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hi everyone for this problem, it reads if 100 and 50 mL of 2.5 times 10 to the negative three molar of hydra bronek acid is mixed with mL of 1.5 times 10 to the negative three molars of nitric acid. What is the ph of the resulting solution? Assume the volumes are additive. So we want to answer the question of the P. H. Of the resulting solution. So let's just take a note at what types of acids we're dealing with here. So we have hydro bronek acid and we also have nitric acid. So these are both strong acids. Okay, and remember that strong acids they dissociate completely. Okay, so t they dissociate completely. And the first thing that we're going to want to do to solve this problem is we're going to want to find the malaria T. Of each acid using the dilution formula. And the dilution formula is M one V one equals M two V two. So let's go ahead and find the malaria T of hydra bronek acid first. Okay, so malaria T of hydro bronek acid. And let's write down what we know based off of what is given in the problem. We know that M one is equal to 2.5 times 10 to the negative three moller V one is equal mL and V two is going to be the combination of both. Okay, so we have 100 and 50 mL plus mL because the volumes are additive. So V two is equal to 200 mL. So M two equals what that is going to be the mill arat E of the hydroponic acid. So we need to rearrange our a dilution formula so that we're solving for M. Two. So when we rearrange this formula, what it now becomes is M. Two is equal to M. One times V one over V. Two. Okay, so let's go ahead and plug in the values based off of what is given. So we have M. One which is 2.5 times 10 to the negative three moller V one is 100 and 50 mL and this is all over V. Two which is 200 mL. So the malaria T. M two, once we solve is 1.8 75 times 10 to the -3 molar. Okay, and with this we said that hydra bronek acid is a strong acid. Okay, so with it being a strong acid, it is going to disassociate completely. And when we write this out in terms of a reaction, what this becomes is our hydro bronek acid plus water is going to yield hydro ni um ion plus the bromide and ion. So what this means then is the concentration of hydro bronek acid equals the concentration of hydro knee um ion because this is a strong acid. Okay, so that means then the concentration of hydrochloric acid is 1.875 times 10 to the negative three molar because it's a strong acid. All right, so we know then the first part. Okay, so we know that are we know what our concentration of hydro bronek acid is? Or are malaria T of hydrochloric acid. So now let's go ahead and solve for the mill arat E of nitric acid. Mhm. So, we're going to go ahead and do the same thing. So R M one is equal to 1.5 times 10 to the negative three moller. V one is equal to mL. V two is going to be the sum of both volumes. So same thing. We have 100 and 50 mL plus 50 mL gives us a V two of 200 mL and M two equals what that's going to be our polarity of the nitric acid. So, we're going to do the same thing that we just did up above. So, we solve for M. Two. So M two is equal to M one times V one over V two. So we're gonna plug in what we know. So, we have M one is 1.5 times 10 to the negative three molar times V one is mL over V. Two, which is 200 mL. So when we solve for M. Two, What we get is 3.75 times 10 to the -4 moller. And just like hydra bronek acid. Nitric acid is a strong acid, which means it's going to disassociate completely. Okay. And so in terms of a reaction when we have nitric acid react with water, it's going to produce hydro ni um ions plus the nitrate and ion. So this means that the concentration of nitric acid is going to equal the concentration of hydro knee um ion, which we just saw four and we said is 3.75 times 10 to the negative four moller. So now we know what the concentration is for the nitric acid. So we used the we found we found the mill arat E of each acid using the dilution formula and we found the mill arat E of the hydro ni um ions using the mill arat ease of each acid. Now the last thing that we need to do is find the ph of the solution. Okay, so for the ph of the solution, the total mill arat E. Of the hydrogen ions in the solution is going to equal the sum of these two polarities that we just solved for. Okay, so our total mill arat E is equal to 1.875 times 10 to the negative three moller plus 3.75 times 10 to the negative four molar. So the total polarity is equal to 2.25 times 10 to the negative three molar. And remember the P. H of a solution is equal to negative log the concentration of hydro ni um ions. Okay and the total mill arat E. Is going to equal that number for us for the total number of hydro knee um ions. So P. H. Is equal to negative log of our hydro ion concentration, which is the total mill arat E. And we said that value is 2.25 times 10 to the negative three molar. So we can now solve for P. H. And what we're going to get for the final P. H. is it is equal to 2.65. Okay, so that is going to be it for this problem. I hope this was helpful.

Calculate the pH of solutions prepared by: (d) Mixing 100.0 mL of 2.0 * 10-3 M HCl and 400.0 mL of 1.0 * 10-3 M HClO4. (Assume that volumes are additive.)

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Key Concepts

pH Scale

Strong Acids

Dilution and Volume Addition