Determine the [OH-] concentration in a 0.344 M Ba(OH)2 solution. 0.463 M 0.688 M 0.334 M...

Question

Question

Answer 1

Answer #1

Given that concentration of Ba(OH)₂ is 0.344M.

Now, Ba(OH)_{2 ^—>} Ba²⁺ + 2 OH ^-

Hence, 1M of Ba(OH)₂ Produce 1M of Ba2+ and 2M of OH-

So, in 0.344M of Ba(OH)₂ there will be 2×0.344M = 0.688M of OH-

Therefore [OH^-] = 0.688M

Answer 2

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