Calculate the pH of 100.0 mL of 0.30 M NH3 before and after the a...

Question

Calculate the pH of 100.0 mL of 0.30 M NH3 before and after the addition of 4.0 g of NH4NO3, and account for the change. Assume that the volume remains constant.

Accepted Answer

Hello. In this problem, we are asked to determine the ph of 100 and 50 mL of a point for Mueller Piratey in solution were asked what is the ph after edition of five g of pyrenean hydrogen chloride were told to assume that the addition of the pie radium hydrochloride does not affect the volume. Let's begin then by finding the ph then with just our base present. So that pirating and we'll begin by creating a nice table. So we'll write our reaction then between our base and its reaction with water to inform the conjugate acid of our base and hydroxide ions. So it is the base. So it will accept a proton from water. So initial change in equilibrium. So initially we have 0.40 moller Our poverty will ignore water because it's a pure liquid and we initially have none of our conjugate acid or hydroxide. The changes minus X plus X and plus X. The initial change is equal to the equilibrium and so our equilibrium constant. So our base association constant will need to find in a table. So usually at the back of our textbooks at 1.79 times 10 to minus nine the sun is equal to our product concentrations. But by our reacting concentration. Using the values from the ice table, we then want to determine whether or not we can simplify this. So eliminate the minus X. So we'll check are simplifying assumption. So to do that we take our Initial concentration of the pirate 18 divided by our equilibrium constant. And this works out to 2.23 Times to the eight which is much larger than 500. So it's safe for us to then simplify the calculation and eliminate the -X. We get 1.79 times 7 -9. Then is equal to X word over 0.40, solve for X. Multiply both sides by 0.4 and then take the square root. Next, then works out to 2.68 Times 10 to -5 smaller. This is equal to then the concentration of our hydroxide ions shown in our rice table. So we can find the P O. H. The P O. H. Then is the negative log of our hydroxide iron concentration. And so this works out to then 4. ph. Then we take 14 minus The P. O. H. This works out to 9.4 - seven. So the ph of our based solution. The pirating is 9.427 which makes sense. It's greater than neutral. Um since this is the base we expect PHP greater than seven. Next then we will find the ph with the addition of five g of our Canadian hydrogen chloride. So we'll begin by finding the initial concentration the palladium hydrogen chloride that we can then make use of in our rice table. So we have five g and we're told that didn't impact the volume. So still have 100 50 mL of our solution will convert from middle leaders. two leaders and will convert from master moles. That's one roll of Hi radium hydrochloride has amassed 114 .553g. And in every one mole I'm done Caribbean, Hydrogen Chloride. We have one mole than of our poverty and ion this works out to then Your . Moller. So we see then that our unit milliliters cancels our units of grams cancels moles of piratey um sergeant Floyd cancels and we're left with then molds of our harry deny on her later. Well now make our ice table. Oops. So our reaction again the hydro sis of our base from why don't you get acid and hydroxide islands initial change in equilibrium. So initially then we have 0.40 of the pirating. Will ignore water because it's pure liquid. We now initially have some of the acid component. So we have a concentration of 0.29. And initially no hydroxide changes minus X plus X. And plus X. You have 0.40 - 0.29 Plus X&X. So based on what we found for the simplifying assumption with the previous calculation, you can simplify our calculation. So the X. can be small relative to 0.40. So we can Estimate this is just .40 and the X will be small relative to 0.29. So this then can be approximated at 0.29. Our equilibrium cost expression. Then again we have the concentrations our products over that of our reactant because then is equal to 0.29 times X, divided by 0.40, which is equal to 1.79 times 10 minus sign, solving for X. Get X is equal to 2. times 10 to -9 Holer which is equal to our hydroxide ion concentration. So the P. O. H. People to the negative log of our hydroxide ion concentration. So are P. O. H. works out to eight 609 and our p is equal to 14 minus the P. O. H. This works out then to 5.39. So the ph of our solution with the addition of 5g of pirating hydrant cord is 5.39. We see that the ph decreased upon the addition of the um common ion which was our um acid component, the pirate Ian ion. Thanks for watching. Hope this helped

Calculate the pH of 100.0 mL of 0.30 M NH3 before and after the addition of 4.0 g of NH4NO3, and account for the change. Assume that the volume remains constant.

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