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3 years ago

How many grams of copper (II) nitrate would be produced from 0.80 g of copper metal reacting with excess nitric acid?

Chemistry

1 answer:

zaharov [31]3 years ago

7 0

Answer:

$m_{Cu(NO_3)_2}=2.36 gCu(NO_3)_2$

Explanation:

Hello!

In this case, since the chemical reaction between copper and nitric acid is:

$2HNO_3+Cu\rightarrow Cu(NO_3)_2+H_2$

By starting with 0.80 g of copper metal (molar mass = 63.54 g/mol) and considering the 1:1 mole ratio between copper and copper (II) nitrate (molar mass = 187.56 g/mol) we can compute that mass via stoichiometry as shown below:

$m_{Cu(NO_3)_2}=0.80gCu*\frac{1molCu}{63.54gCu} *\frac{1molCu(NO_3)_2}{1molCu} *\frac{187.56gCu(NO_3)_2}{1molCu(NO_3)_2} \\\\m_{Cu(NO_3)_2}=2.36 gCu(NO_3)_2$

However, the real reaction between copper and nitric acid releases nitrogen oxide, yet it does not modify the calculations since the 1:1 mole ratio is still there:

$4HNO_3+Cu\rightarrow Cu(NO_3)_2+2H_2O+2NO_2$

Best regards!

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The pH of a saturated solution of a metal hydroxide MOH is 10.15. Calculate the Ksp for this compound.

Airida [17]

Answer:

$Ksp=2.00x10^{-8}$

Explanation:

Hello!

In this case, since the pH of the given metal is 10.15, we can compute the pOH as shown below:

$pOH=14-pH=14-10.15=3.85$

Now, we compute the concentration of hydroxyl ions in solution:

$[OH^-]=10^{-pOH}=10^{-3.95}=1.41x10^{-4}M$

Now, since this hydroxide has the form MOH, we infer the concentration of OH- equals the concentration of M^+ at equilibrium, assuming the following ionization reaction:

$MOH(s)\rightarrow M^+(aq)+OH^-(aq)$

Whose equilibrium expression is:

$Ksp=[M^+][OH^-]$

Therefore, the Ksp for the saturated solution turns out:

$Ksp=1.41x10^{-4}*1.41x10^{-4}\\\\Ksp=2.00x10^{-8}$

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