4 years ago

Estimate the solubility of m(oh)2 in a solution buffered at ph = 7.0, 10.0, and 14.0.

1 answer:

4 0

When M(OH)2 dissolves we have M(OH)2 which produces M2+ and 2OHâ’ pH + pOH=14 At ph =7; we have 7+pOH=14 pOH=14â’7 = 7 Then [OHâ’]=10^(â’pOH) [OH-] = 10^(-7) = 1* 10^(-7) At ph = 10. We have, pOH = 4. And [OH-] = 10^(-4) = 1 * 10^(-4) Finally ph = 14. We have, pOH = 0 And then [OH-] = 10^(-0) -----anything raised to zero power is 1, but (-0)... So [OH-] = 1

You might be interested in

I need help with this ASAP PLEASE

Artemon [7]

439.3 g CO2

Explanation:

First find the # of moles of CO2 that results from the combustion of 3.327 mol C3H6:

3.227 mol C3H6 × (6 mol CO2/2 mol C3H6)

= 9.981 mol CO2

Use the molar mass of CO2 to determine the # of grams of CO2:

9.981 mol CO2 x (44.01 g CO2/1 mol CO2)

= 439.3 g CO2

5 0