Buffer Solutions

I need to calculate the pkb, the first pkb is a negative log of kb, so it will be a negative log of 1.8 times ten to the power of negative five and we have seen that value, so this is four point seven four four seven three now the pka plus the pkb these two add up to 14. so to calculate the pka it will be 14 minus the pkb, so 14 minus 4.74473 we have a pka value of 9.25527, so since we have more base than acid, we know that the answer is that the pH of the solution has to be greater than the pka, it has to be greater than 9.25527.

Let's round that up to 9.255, so let's go ahead and calculate the pH of the solution, but before we do that, notice that the ratio of base to acid is ten to one, one point five divided by point fifteen is ten. As long as the ratio is ten to one or one to ten, we know that the ph will differ from the pka. by one unit and since we have more base and acid, a unit greater than this number will be 10.255, therefore, conceptually we know that the pH of this solution must be this value and we are going to confirm it by calculation, so we should obtain 10.255 so let's write the

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equation ph equals pka ​​plus log base over acid, the pka is 9.255 and then we can add 2 7 as well to get a more accurate result and then add log now as mentioned before. units of molarity or it could be in moles, both units will cancel out, so let's enter the moles.

We have 1.5 moles over the 15 mole point, so we can cross out these two units. Now let's enter this into our calculator nine point two five. five two seven plus log one point five over point fifteen this will give you ten point two five five two seven so that's the pH of the solution so we can see why it's one unit higher because the ratio of base to acid is ten to one number three calculates the pH of a solution containing 15 grams of hydrofluoric acid and 21 grams of sodium fluoride and 750 milliliters of solution, so once again we have a

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solution, but this time we are given the grams of the acid and the base before I had the molarity and during the second problem we had the moles for this one all we need to do is convert grams to moles and then we could use the buffer equation so let's do that first to have 15 grams of hf to convert to moles . we need to know the molar mass the molar mass of hydrogen is about one for fluorine is 19. so this is about 20 grams per mole so one mole of hf has a mass of about 20 grams so we can cross out the unit grams of hf will be 15 divided by 20, which is 0.75, so that's how many moles of acid we have in this solution.

Let's start with 21 grams of the naf base and convert it to moles. Sodium has an atomic mass of about 23 and fluorine is 19 23 plus 19 is 42. so one mole of naf has a mass of about 42 grams 21 divided by 42 is 0.5 so we get 0.5 moles of sodium fluoride so now that we have the moles of the acid and the base now we can calculate the ph but first let's calculate the pka the pka which is the negative log of ka and ka is that number, so we have the negative log of seven point two times ten to the power of negative four and that will be three point 3.1427, so now that we have to let's use the henderson-hasselbach equation to get the ph of the solution, so the pka is 3.1427 plus the log of the base, by the way, the base will be greater or less than the pka, so notice that we have more of the acid than the base, so because we have more acid than base, the ph will be less than the pka, so the ph must be less than 3.1427, the solution is acidic, so now let's go ahead and continue with this equation so the base is 0.5 moles, the acid is 0.75 moles, so let's plug this into 3.1427 plus log 0.5 over 0.75, this It will be approximately 2.97, which is less than 3.14, so that is the pH of the solution, so every time it has more acid component than the base component, the pH will be less than pka number four, which is the pka of an unknown weak acid if the pH of the solution was measured to be 5.62 when h a is 0.45 and a minus is 0.85 then we want to calculate the pka if we know the ph and concentration of the weak acid and the weak conjugate base so Let's start with the buffer equation.

The pH is equal to pka plus log base over the acid. Now what we need to do is isolate pka, so I'm going to take this term move it to the other side so that it's positive on the right side it's going to be negative on the left side now I'm going to change the left side of the equation with the right side from the equation so we can say that the pka of the unknown weak acid will be the ph of the solution minus the log of the base over the acid, this is how we can calculate the pka of an unknown weak acid, all we need to know is the pH of the solution and the amount of acid and base that we have dissolved in the solution, so the pH is 5.62, the base is the concentration of a minus, so it is 0.85, the acid is the concentration of ha, so it is 0.45, so this will be 5.34, so this is the pka of the unknown acid now let's see if our answer makes sense, so whenever the base has a higher concentration in a amount than the acid, we know that the pH is going to be higher than the pka, that is the situation here.

The ph is 5.62, the pka is 5.34, so we can clearly see that the ph is greater than the pka, so this answer makes sense.