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2 years ago

What is the pH of a 0.0100 M sodium benzoate solution? Kb (C7H5O2-, ) = 1.5 x 10^-10

Chemistry

1 answer:

77julia77 [94]2 years ago

7 0

The pH of a 0.0100 M sodium benzoate solution with a Kb of 1.5 × 10-¹⁰ is 5.91 (option B).

<h3>How to calculate pH?</h3>

The pH of a solution in chemistry refers to the notation expressing the acidity or alkalinity of a solution.

The pH is measured on a logarithmic scale of which 7 is neutral, lower values are more acid and higher values more alkaline.

The pH of a solution can be calculated using the following formula;

pH = - log {H+}

pH = - log {0.0100}

pH = 2

Therefore, the pH of a 0.0100 M sodium benzoate solution with a Kb of 1.5 × 10-¹⁰ is 5.91.

Learn more about pH at: brainly.com/question/11300720

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Read 2 more answers

Two different bromide solutions are mixed with each other: Solution 1 is an aqueous solution of 4.85 g aluminum bromidein 150. m

erma4kov [3.2K]

Answer:

M=0.380 M.

Explanation:

Hello there!

In this case, given those two solutions of aluminum bromide and zinc bromide, it is firstly necessary to compute the moles of bromide ions in each solution as shown below:

$n_{Br^-}^{in\ AlBr_3}=4.85 gAlBr_3*\frac{1molAlBr_3}{266.69gAlBr_3}*\frac{3molBr^-}{1molAlBr_3} =0.05456molBr^-\\\\n_{Br^-}^{in\ ZnBr_2}=7.75gZnBr_2*\frac{1molZnBr_2}{225.22gZnBr_2}*\frac{2molBr^-}{1molZnBr_2} =0.06882molBr^-$

Now, we compute the total moles of bromide:

$n_{Br^-}=0.05456mol+0.06882mol\\\\n_{Br^-}=0.12338mol$

Then, the total volume in liters:

$150mL+175mL=325mL*\frac{1L}{1000mL} \\\\=0.325L$

Therefore, the concentration of total bromide is:

$M=\frac{0.12338mol}{0.325L}\\\\M=0.380M$

Best regards!

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