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andreyandreev [35.5K]
3 years ago
5

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

Chemistry
1 answer:
kherson [118]3 years ago
4 0
<span>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</span>
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Answer:

129.73 g of CaBr₂

Explanation:

We'll begin by writing the balanced equation for the reaction. This is illustrated below:

CaBr₂ + 2KOH –> Ca(OH)₂ + 2KBr

Next, we shall determine the mass of CaBr₂ that reacted and the mass of Ca(OH)₂ produced from the balanced equation. This can be obtained as follow:

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SUMMARY :

From the balanced equation above,

200 g of CaBr₂ reacted to produce 74 g of Ca(OH)₂.

Finally, we shall determine the mass of CaBr₂ that react when 48 g of Ca(OH)₂ were produced. This can be obtained as follow:

From the balanced equation above,

200 g of CaBr₂ reacted to produce 74 g of Ca(OH)₂.

Therefore, Xg of CaBr₂ will react to produce 48 g of Ca(OH)₂ i.e

Xg of CaBr₂ = (200 × 48)/74

Xg of CaBr₂ = 129.73 g

Thus, 129.73 g of CaBr₂ were consumed.

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