Duplicate Question
The question on this page has been marked as a duplicate question.
Original Question
You need to prepare 100.0 mL of a pH=4.00 buffer solution using 0.100 M benzoic acid (pKa = 4.20) and 0.240 M sodium benzoate....Asked by Adam
                You need to prepare 100.0 mL of a pH=4.00 buffer solution using 0.100 M benzoic acid (pKa = 4.20) and 0.180 M sodium benzoate
How much of each solution should be mixed to prepare this buffer?
            
        How much of each solution should be mixed to prepare this buffer?
Answers
                    Answered by
            DrBob222
            
    Let x = mL benzoate
Then 100-x = mL acid
--------------------
millimols base = x*0.180
mmoles acid = (100-x)*0.1
4.00 = 4.20 + log(0.180x)/[(100-x)*0.1]
Solve for x and 100-x
I get 26 something for base and about 74 for acid. You should do it more accurately.
    
Then 100-x = mL acid
--------------------
millimols base = x*0.180
mmoles acid = (100-x)*0.1
4.00 = 4.20 + log(0.180x)/[(100-x)*0.1]
Solve for x and 100-x
I get 26 something for base and about 74 for acid. You should do it more accurately.
                    Answered by
            Hassan
            
    I need your helping for assigment of account
    
                    Answered by
            Hassan
            
    Ineed helping full for account
    
                                                    There are no AI answers yet. The ability to request AI answers is coming soon!
                                            
                Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.