Calculate the pH of the solution that results from each mixture. b. 125.0 mL of 0.10 M NH₃ with 250.0 mL of 0.10 M NH₄Cl

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hey everyone, we're asked to consider a 250 millimeter of a buffer solution that is 0.396 Mohler in methyl amine and 0. moller in methyl ammonium bromide, calculate initial and final ph after adding 0.0 to 50 mol of potassium hydroxide. Looking at our problem since we do have a weak base, we have to use the following formula which would be our Henderson Hasselbach equation. But in terms of our P O H and r p K B. So we have P O H equals P K B plus the log of our conjugate acid divided by our weak base. And looking at our textbooks are PKB is going to be 3.35. Now let's go ahead and use this formula to calculate our P O. H. To get to our ph So plugging in our values, we have P O h equals 3.35 plus the log of 0.178 which was provided to us Divided by 0.396 which was also provided to us in our questions them. So when we calculate this out, we end up with 3.27. Now to calculate our ph all we need to do is take 14 and subtract 3.27 which will get us to a ph of 10.997. Now let's go ahead and calculate our final ph so adding a base consumes the conjugate acid and produces more weak base. So first let's go ahead and calculate our moles. So for our mole of methyl amine, We have 0.396 Mohler. And we're going to multiply this by 0.25 L and this will get us to 0.099 mol. Our molar itty is equivalent to moles over liters. So that is why our leaders cancel out. And we're left with moles. Now looking at our mole of methyl ammonium bromide Again, we're going to take 0.178 moller And multiply this by 0.25 l And this will get us to a value of 0.0445 mol. Next, we're going to calculate our moles after potassium hydroxide edition so far mole of methamphetamine, We're going to take that value of 0.099 mol And we're going to add 0.0-50 mol of our potassium hydroxide. This will get us to a total of 0.1-4 mol. Next calculating our mole of methyl ammonium bromide, We're going to take that value of 0.0445 mol And we're going to add 0.0-50 mole, which will get us to a value of 0.0195 mol. Now to calculate our final ph to calculate our final ph we're going to use the Henderson Hasselbach equation that we wrote above. So plugging in our values we have our p O H equals 3.35 plus the log of 0.195 which is our conjugate acid Divided by 0.124, which was our weak base. And we can use our molds directly since our volumes cancel out. So when we calculate this out, we end up with a p O. H of 2.5466. Now to calculate our final ph again, we're going to take 14 and subtract 2.5466. This will get us to a ph of 11.453. Now, I hope this made sense. And let us know if you have any questions.

Calculate the pH of the solution that results from each mixture. b. 125.0 mL of 0.10 M NH₃ with 250.0 mL of 0.10 M NH₄Cl

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Key Concepts

Acid-Base Chemistry

Henderson-Hasselbalch Equation

Dilution and Concentration

Calculate the pH of the solution that results from each mixture. b. 125.0 mL of 0.10 M NH3 with 250.0 mL of 0.10 M NH4Cl

Verified Solution

Key Concepts

Acid-Base Chemistry

Henderson-Hasselbalch Equation

Dilution and Concentration

Calculate the pH of the solution that results from each mixture. b. 125.0 mL of 0.10 M NH₃ with 250.0 mL of 0.10 M NH₄Cl