Blood is buffered by carbonic acid and the bicarbonate ion. Normal blood plasma is 0.024 M in HCO₃^- and 0.0012 M H₂CO₃ (pK_a1 for H₂CO₃ at body temperature is 6.1).

c. Given the volume from part (b), what mass of NaOH can be neutralized before the pH rises above 7.8?

Video transcript

Hello. In this problem we are told that urine is buffered by hydrogen phosphate and hydrogen phosphate. If six liters of phosphate buffer has 0.10 Mueller hydrogen phosphate and 0.10 Mueller di hydrogen phosphate. What is the mass of potassium hydroxide that can be neutralized by the buffer before the ph rises above 8.10. We're giving the K. Value for di hydrogen phosphate. So since we have a buffer we can make use of the Henderson Hasselbach equation which says the ph is equal to the P. K. Plus the log of the concentration of our base component. So that will be hydrogen phosphate to that of our acid component which will be di hydrogen phosphate. We can find the P. K. It's equal to the negative log of K. A. This negative log then of 6.2 times 10 to the -8. This works out to then 7.208. We can make use of the Henderson has a back equation to find the ratio of the concentration of our base component to that of our acid component. Subtract P. K. From both sides and then take the inverse log We then get the ratio of our concentration is equal to to the ph minus the P. K. A. The ph is 8.10 R. P.K. This works out then to 7.798. The concentration of our base component Is equal to 7.7 98 times the concentration of our acid component. We'll now generate a table that describes the reaction of they are added base with a buffer. We have the acid component reacting with the hydroxide perform hydrogen phosphate and water. We have initial change and final. So initially the concentration of dye hydrant phosphate is 0.10 moller. The amount of hydroxide iron that can be neutralized by the buffer will call y and r initial amount of the hydro phosphate is 0.10 moller will ignore water. Since it's a pure liquid, we're gonna assume that all the hydroxide reacts. So that would be minus Y. Based on the story geometry for every one mole of hydroxide that reacts an equal amount of our acid component will react and an equal more amount of our base component will be formed. We're going to combine the initial and the change to come up with the final and then plug these final concentrations into our equation given above the table We have the concentration of our base 0.010 plus y is equal to 7.798 times the concentration of our acid component will simplify the right hand side. We'll move our Y components together and those without are variable together and then solve for y. This works out to 0. Moller. This is equal to then the concentration of hydroxide ions that can be neutralized by the buffer solution. We're putting the hydroxide ions into six liters of buffer. Will you make use of the concentration of hydroxide iron to go from volume to moles And for every one mole of hydroxide, We had one mole of potassium hydroxide. Thank you of miller mass of potassium hydroxide. One mole of potassium hydroxide has a massive 56 grams. Our leaders canceled most of hydroxide cancels moles of potassium hydroxide cancels and we're left with grams of potassium hydroxide. This works out to 29 g of potassium hydroxide. This is the massive potassium hydroxide that can be neutralized by the buffer before the ph rises above 8.10. This answer corresponds to D. Thanks for watching. Hope. This helped.