# The value of Ksp for Ni(OH)2(aq) is 6.0x10^-16. What is the molar concentration of nickel ions in a saturated pH=7.0 buffer solution? Assume the buffer is sufficiently strong that the pH does not change during equilibration.

## The value of Ksp for Ni(OH)2(aq) is 6.0x10^-16. What is the molar concentration of nickel ions in a saturated pH=7.0 buffer solution? Assume the buffer is sufficiently strong that the pH does not change during equilibration.

Nov 17, 2016

It is $6.0 \times {10}^{-} 2$.

#### Explanation:

There is a big hint in the name: Ksp is the equilibrium solubility product constant. In other words, you are going to take the solubility product, that is, you are going to multiply two solubilities, to get the equilibrium equation!

The equilibrium equation is:

$N i {\left(O H\right)}_{2} r i g h t \le f t h a r p \infty n s N {i}^{2 +} + 2 O {H}^{-}$

In this case we are going to multiply the solubility of the $N {i}^{2 +}$ ion by the concentration of the hydroxide ion:

${K}_{s p} = \left[N {i}^{2 +}\right] \cdot {\left[O {H}^{-}\right]}^{2}$

Remember that, as with any equilibrium equation, we always raise the concentration of the chemical to its coefficient in the balanced chemical equation.

So that means that

$6.0 \times {10}^{- 16} = \left[N {i}^{2 +}\right] \cdot {\left[O {H}^{-}\right]}^{2}$

Now at pH 7.0 the

$\left[O {H}^{-}\right] = 1 \times {10}^{- 7}$

So...

$6.0 \times {10}^{- 16} = \left[N {i}^{2 +}\right] \cdot {\left[1 \times {10}^{- 7}\right]}^{2}$

Dividing both sides by ${\left[1 \times {10}^{- 7}\right]}^{2}$

We get

$\left[N {i}^{2 +}\right] = \frac{6.0 \times {10}^{- 16}}{1 \times {10}^{- 7}} ^ 2 = 6.0 \times {10}^{- 2}$