The concentration of C6H12O6 may be represented as (C6H12O6) {C6H12O6} [C6H12O6]
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In this case, since the concept of solution concentration is majorly used in terms of moles per liters of solution or molarity (M), it is also possible to represent this chemical unit by using squared brackets, [ ].
In such a way, when focused on the concentration of glucose, C6H12O6, we can use:
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The dissociation reaction for Benzoic Acid is the following:
C₆H₅COOH + H₂O ⇄ C₆H₅COO⁻ + H₃O⁺
The Ka expression is the following and we clear for the concentration of H₃O⁺(X) assuming that the dissociation is little so we can rule it out in the denominator of the equation:
![Ka= \frac{[C_6H_5COO^{-}]*[H_3O^{+}] }{[C_6H_5COOH]}=\frac{X*X }{0,40 -X} (assume: 0,40-X\approx0,40)\\ \\ X= \sqrt{0,40*6,50*10^{-5} }=0.00510M \\ \\ pH=-log([H_3O^{+}]=2,29](https://tex.z-dn.net/?f=Ka%3D%20%5Cfrac%7B%5BC_6H_5COO%5E%7B-%7D%5D%2A%5BH_3O%5E%7B%2B%7D%5D%20%7D%7B%5BC_6H_5COOH%5D%7D%3D%5Cfrac%7BX%2AX%20%7D%7B0%2C40%20-X%7D%20%20%28assume%3A%200%2C40-X%5Capprox0%2C40%29%5C%5C%20%20%5C%5C%20X%3D%20%5Csqrt%7B0%2C40%2A6%2C50%2A10%5E%7B-5%7D%20%7D%3D0.00510M%20%5C%5C%20%20%5C%5C%20pH%3D-log%28%5BH_3O%5E%7B%2B%7D%5D%3D2%2C29%20%20%20)
So, the pH of this Benzoic Acid solution is 2,29
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The dissociation reaction for Benzoic Acid is the following:
C₆H₅COOH + H₂O ⇄ C₆H₅COO⁻ + H₃O⁺
The Ka expression is the following and we clear for the concentration of H₃O⁺(X) assuming that the dissociation is little so we can rule it out in the denominator of the equation:
So, the pH of this Benzoic Acid solution is 2,29
Have a nice day!