B) Consider the mixing of sodium hypobromite (NaBrO) into...

b) Consider the mixing of sodium hypobromite (NaBrO) into 2.00 L of 0.25 M hypobromous acid (HBrO) to form a buffer solution (Ka of HBrO = 2.3 x 10^-9)? i. Assuming that no volume change occurs when the NaBrO is added, Calculate the number of moles of NaBrO need to be added into the solution to form a buffer solution with pH of 8.20 ii. Calculate the final pH if 10 mL of 0.5 M NaOH is added into the solution.

Transcript

00:01 Hello everyone, let's start the question.

00:03 In this question particularly we're talking about the formation of a buffer which is formed by mixing a solution of nabro into 0 .25 molar which is the molarity for hbro solution and they are forming a buffer together.

00:20 So first of all we have the given information that we have the volume provided which is 200 l.

00:30 And we have the concentration of hbro which is equal to 0 .25 molars.

00:46 Also we have the value of dissociation constant of acids given as 2 .3 multiplied by 10 raised to the power negative 9.

00:58 We have to calculate the p .a.

01:00 So we know that pca is actually equal to negative logarithm of the ka value and if you substitute it here we will obtain the value of pca for this particular solution as 8 .64.

01:16 This is our value for pkk.

01:18 Now let's come to the main part, that is the first part of the question, in which we have to calculate the number of moles of any bro needed.

01:31 Number of moles we need to calculate for n -a -b -r -o which is required to make buffer, to make buffer, this is to be calculated for which we have the ph value provided as 8 .20.

01:57 Now let's use the given information and proceed.

02:05 By using the henderson equation using henderson equation for which we have the relation that ph is equal to the p -k -a value added to logarithm of the concentration of salt divided by the concentration of acid.

02:36 If we substitute the values here, the value of ph is 8 .20.

02:43 It is equal to the pca value which is 8 .64 as calculated above and we will add the logarithm of salt which is nabro, the concentration of salt divided by the concentration of acid which is hbro...

Question

Please give Ace some feedback