A buffer contains 0.10 mol of acetic acid and 0.13 mol of sodium acetate in 1.00 L. (b) What is the pH of the buffer after the addition of 0.02 mol of KOH?

Video transcript

Hey everyone. So you were asked to consider a solution as 25 g of acetic acid and 15 g of sodium acetate. And a 250 male leader of solution. And we asked calculate the P. H. Using the Henderson Hasselbach equation. Recall that this is gonna be P. H. It was P. K. A. That's the law of the conjugate base by the weak acid. So we need to first calculate the moles of each. We have 25 g of acetic acid. And in one bowl of acetic acid we have the molar mass. It's gonna be too Times 12.011g. Last four. That was 1.008 g. Last two I was 15.999 g. And we get 60.06 g. We're gonna get 0.416 moles of acetic acid. They were 15 g a sodium acetate. And in one more a sodium acetate at the molar mass. We're gonna get 22 .990 g. Last two 12.011 g. Last three I was 1.008 g. Last two I was 15.999 g. We're gonna get 82 .04 g. This will give us 0.183 balls of sodium acetate in their P. K. A. It's 4.75. So now we can plug everything into the equation and solve for the ph we're gonna have ph Eagles. plus a law Of 0. malls. What about 0.416 malls? Well, for the ph We're gonna get 4.39. Thanks for watching my video. And I hope I was helpful.