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ch4aika [34]
2 years ago
9

Will give brainiest

Chemistry
1 answer:
zimovet [89]2 years ago
3 0
MM=97.10 that’s the answer
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3 years ago
A 25.0 ml sample of 0.150 m benzoic acid is titrated with a 0.150 m naoh solution. what is the ph at the equivalence point? the
Ad libitum [116K]

Solution:

At the equivalence point, moles NaOH = moles benzoic acid  

HA + NaOH ==> NaA + H2O where HA is benzoic acid  

At the equivalence point, all the benzoic acid ==> sodium benzoate  

A^- + H2O ==> HA + OH- (again, A^- is the benzoate anion and HA is the weak acid benzoic acid)  

Kb for benzoate = 1x10^-14/4.5x10^-4 = 2.22x10^-11  

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x^2 = 3.33x10^-12  

x = 1.8x10^-6 = [OH-]  

pOH = -log [OH-] = 5.74  

pH = 14 - pOH = 8.26  


5 0
3 years ago
Read 2 more answers
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