The electrolysis of 282 g of a 40% solution of copper (II) nitrate was stopped after, when the mass of the solution decreased

The electrolysis of 282 g of a 40% solution of copper (II) nitrate was stopped after, when the mass of the solution decreased by 32 g. 140 g of a 40% sodium hydroxide solution was added to the resulting solution. Determine the mass fraction of alkali in the resulting solution.

Let us write the equation for electrolysis of an aqueous solution of copper (II) nitrate:
2Cu (NO3) 2 + 2H2O → (electrolysis) 2Cu + O2 + 4HNO3
The decrease in the mass of the solution was due to the release of copper at the cathode and oxygen at the anode.
Let’s check whether copper (II) nitrate remained in the solution after the end of electrolysis (when Cu (NO3) 2 completely reacts, water electrolysis will begin).
Let’s find the mass and amount of the substance of the initial copper (II) sulfate:
m (Cu (NO3) 2) ref. = m (Cu (NO3) 2) p – pa ∙ ω (Cu (NO3) 2) = 282 g ∙ 0.4 = 112.8 g
n (Cu (NO3) 2) ref. = m (Cu (NO3) 2) ref. / M (Cu (NO3) 2) = 112.8g / 189g / mol = 0.6 mol.
If all Cu (NO3) 2 is consumed, then according to the electrolysis equation the mass of the formed copper will be 0.6 mol ∙ 64 g / mol = 38.4 g, which already exceeds the sum of the masses of copper and oxygen (32 g) released from the solution. Therefore, after electrolysis, Cu (NO3) 2 remained in the solution.
The added sodium hydroxide reacts with the remaining Cu (NO3) 2 and the resulting nitric acid:
Cu (NO3) 2+ 2NaOH → Cu (OH) 2 ↓ + 2NaNO3 (1)
HNO3 + NaOH → Na2SO4 + Н2О (2)
Let the amount of the substance of the formed oxygen n (O2) = hmol. Then the amount of the substance of the formed copper n (Cu) = 2x mol. m (O2) = 32x (g), am (O2) = 64 ∙ 2x = 128x (g). By the problem statement: m (O2) + m (O2) = 32.
32x + 128x = 32
x = 0.2 (mol)
Let us find the amount of copper (II) nitrate substance subjected to electrolysis:
n (Cu (NO3) 2) reac. = n (Cu) = 2x mol = 2 ∙ 0.2 mol = 0.4 mol.
Let’s find the amount of copper (II) nitrate substance remaining in the solution:
n (Cu (NO3) 2) rest = n (Cu (NO3) 2) ref. – n (Cu (NO3) 2) reac. = 0.6 mol – 0.4 mol = 0.2 mol.
Let’s find the amount of the substance of the formed nitric acid:
n (HNO3) = 2 ∙ n (CuSO4) reac. = 2 ∙ 0.4 mol = 0.8 mol
Determine the mass and amount of the substance of the initial sodium hydroxide solution:
m (NaOH (out.)) in islands = m (NaOH (out.)) solution ∙ ω (NaOH) = 140g ∙ 0.4 = 56g
n (NaOH (ex.)) = m (NaOH (ex.)) in islands / M (NaOH) = 56g / 40 g / mol = 1.4 mol.
Determine the amount of substance and the mass of sodium hydroxide remaining in the solution:
n (NaOH) reaction 1 = 2 ∙ n (CuSO4) rest. = 2 ∙ 0.2 mol = 0.4 mol.
n (NaOH) reac. 2 = n (HNO3) = 0.8 mol.
n (NaOH) rest. = n (NaOH) ref. –n (NaOH) reaction 1 – n (NaOH) reaction 2 = 1.4 mol – 0.4 mol – 0.8 mol = 0.2 mol
m (NaOH) rest. = n (NaOH) rest. ∙ M (NaOH) = 0.2 mol ∙ 40 g / mol = 8 g.
Find the mass of the resulting solution and the mass fraction of sodium hydroxide in it:
mkon.r-pa = m (Cu (NO3) 2) solution + m (NaOH (out)) solution – (m (Cu) + m (O2)) – m (Cu (OH) 2) =
= 282g + 140g – 32g – (0.2 mol ∙ 98 g / mol) = 370.4g
ω (NaOH) end solution = m (NaOH) rest / m end solution = 8g / 370.4 g = 0.216 (2.16%).
Answer: ω (NaOH) = 2.16%.

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