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ololo11 [35]
3 years ago
5

What is the [OH-] in a solution with a poh of 3.90

Chemistry
2 answers:
zheka24 [161]3 years ago
8 0

Answer:

7.9433E-3

Explanation:

POH=-log OH

3.90=POH

OH=ANTILOG OF -3.90

=7.9433×10^-3

andrew-mc [135]3 years ago
5 0

<u>Answer:</u> The hydroxide ion concentration is 1.26\times 10^{-4}M

<u>Explanation:</u>

pOH is defined as the negative logarithm of hydroxide ion concentration in the solution.

To calculate the pOH of the reaction, we use the equation:

pOH=-\log[OH^-]

where,

pOH=3.90

Putting values in above equation, we get:

3.90=-\log[OH^-]

[OH^-]=1.26\times 10^{-4}M

Hence, the hydroxide ion concentration is 1.26\times 10^{-4}M

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7 0
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You are given a solid that is a mixture of na2so4 and k2so4.
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Or, in 0.298 g of the filtered mass (0.298×0.411)=0.122g of SO_{4}^{2-} ion is present.        

5 0
3 years ago
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