Suppose that 00200 mol naohs is dissolved in 300 ml of the...

Suppose that 0.0200 mol NaOH(s) is dissolved in 300. mL of the buffer solution that is 0.040 m NaCH3CO2(aq) and 0.080 mCH3COOH(aq) Calculate the pH of the resulting solution and the change in pH.

Transcript

00:01 Hello everyone according to the question i'm going to be writing the ph is equal to pk plus log here it's going to become salt divided by acid in the very next step i'm going to be writing the ph is equal to 4 minus log 1 .8 plus log acid values we are going to be putting here 0 .2 it's going to be 29 multiplied by 2 60 divided by 0 .205 multiplied by 260 but we are getting answered here 3 .895 okay the basically here we need to be finding out we are going to be calculating the initial ph and the final ph after adding a 0 .0200 mole of nudge what is going to be happening initial and we are supposed to be finding out the initial and the final ph right so initially i'm going to be writing over here for the initial one okay for the initial after adding what after adding and a oh h we are getting over here ph is equal to p -k a plus log it's going to become over here salt plus n -a -o -a -0 divided by acid minus an ao h so what i can write it over here ph is equal to 4 minus log 1 .8 plus log here it's going to become of values that is 0 .0754 plus 0 .02 divided by it's going to become over here 0 533 minus 0 .02 after solving what we are getting after solving we are getting the ph as an i can write this way so ph is equal to 4 .201 okay in the very next step i can write for the second one for a second i am going to be writing over here boh is equal to p k b plus log salt divided by base now we are going to be putting those formula again over here we are going to be writing this formula p o h is equal to 4 minus log 5 .6 plus log 0 .0 okay when we are not 0 it's going to become over here 0 .26 7 5 divided by 0 .2975.

03:09 The value which we are getting here is 3 .206.

03:14 Now we are going to be writing the ph is equal to 14 minus p -o -h is equal to 14 minus 3 .206.

03:26 Here it's going to become 10 .794.

03:30 After adding now we are what we can write after adding after adding an a which is going to be come up so p o h is equal to pkb plus log salt minus n a o h divided by base plus an ao h it's going to be come up over here now we are going to be putting those values so when we are going to be putting the value i can write the p o h is equal to four minus log 5 .6 plus log 0 .06955 minus 0 .02 divided by 0 .0775 plus 0 .0 .07735 plus 0 .02...

Question

Suppose that 0.0200 mol NaOH(s) is dissolved in 300. mL of the buffer solution that is 0.040 m NaCH3CO2(aq) and 0.080 mCH3COOH(aq) Calculate the pH of the resulting solution and the change in pH.

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