The Ka of bromoacetic acid is 2.00×10^-3 2nd attempt What...

The K$_a$ of bromoacetic acid is 2.00$\times$10$^{-3}$ 2nd attempt What masses of bromoacetic acid (CH$_2$BrCOOH) and sodium bromoacetate (CH$_2$BrCOONa) are needed to prepare 1.00 L of pH = 1.900 buffer if the total concentration of the two components is 0.200 M? 14.28 g of bromoacetic acid 16.03 g of sodium bromoacetate

Transcript

00:01 So the question says the cao of bromocytic acid is 2 times times 0 .3.

00:07 What masses of promo acetic acid and sodium bromacetate are needed to prepare one liter of ph 1 .9 buffer.

00:15 So we have the acid and its conjugate base to make a buffer with this ph.

00:23 If the total concentration of the components are 0 .2 molar.

00:28 So we need the grams of bromacetic acid and grams of sodium bromacetic acid, or bromacetate, the conjugate base.

00:38 So we know that ph for buffer, ph equals pca plus log, the concentration of the conjugate base over the concentration of the concentration of the conjugate base at x and concentration of the acid as y.

00:55 We already know that the concentration of the buffer, it's given that it's 0 .2, and the volume is 1 liters.

01:02 So the number of moles for the buffer is going to be 0 .2 moles, that's for sure.

01:06 So the ph is 1 .9, and the ka is 2 .2 times 10th of our negative 3.

01:12 All these are given.

01:13 So let's see how we can solve the equation.

01:15 So pkk equals negative log paa, so negative log 2 times 10th of our negative 3 equals 2 .69.

01:24 So, sitting in the equation, 1 .9 equals 2 .69 plus log concentration of the conjugate base over the concentration of the acid.

01:35 So log x over y would equal negative 0 .79.

01:41 So knowing this, we can know that the x over y equals 0 .61...

Question

Please give Ace some feedback