Calculate the ph of a 0.40 m solution of sodium benzoate (nac6h5coo) given that the ka of benzoic acid (c6h5cooh) is 6.50 x 10-5
1 answer:
Hello!
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
Have a nice day!
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!
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