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Stoichiometry and solutionsNational 2020

A student wishes to prepare a solution with a final concentration of $\ce{Na+}$ = 0.50 M and a final concentration of $\ce{HCO3-}$ = 0.10 M by taking some $\ce{NaOH}$ and some trona ( $\ce{Na2CO3.NaHCO3.2H2O}$ ) and diluting with water to a final volume of 1.00 L. How much $\ce{NaOH}$ and trona are required?

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Explanation

draft

Setting up the species balance

Trona, $\ce{Na2CO3.NaHCO3.2H2O}$ , dissolves to give per mole: $3\ \ce{Na+}$ , $1\ \ce{CO3^2-}$ , $1\ \ce{HCO3-}$ .

Let $x$ = mol trona, $y$ = mol $\ce{NaOH}$ added.

Key reaction: $\ce{NaOH}$ is a strong base and will react with the weaker acid form present, $\ce{HCO3-}$ :
$\ce{OH- + HCO3- -> CO3^2- + H2O}$
So each mole of $\ce{NaOH}$ converts one mole of $\ce{HCO3-}$ into $\ce{CO3^2-}$ — it does not stay as free $\ce{OH-}$ or add extra $\ce{HCO3-}$ .

Bicarbonate balance: starting $\ce{HCO3-}$ from trona is $x$ ; after reacting with $y$ mol $\ce{NaOH}$ , the remaining $\ce{HCO3-}$ is
$x - y = 0.10$

Sodium balance: Na⁺ is conserved regardless of the acid–base reaction (it doesn't change oxidation state or leave solution):
$3x + y = 0.50$

Solve the system:
Add the two equations: $x-y=0.10$ and $3x+y=0.50$ →
$4x = 0.60 \Rightarrow x = 0.15$ mol trona, then $y = x-0.10 = 0.05$ mol $\ce{NaOH}$ .

Check: Na⁺ total $=3(0.15)+0.05=0.45+0.05=0.50$ M ✓; HCO3⁻ remaining $=0.15-0.05=0.10$ M ✓.

So 0.05 mol $\ce{NaOH}$ and 0.15 mol trona are required — answer A.

Correct: recognizes that added $\ce{NaOH}$ reacts with $\ce{HCO3-}$ from trona (converting it to $\ce{CO3^2-}$ ), giving the bicarbonate balance $x-y=0.10$ together with the sodium balance $3x+y=0.50$ , solved to $x=0.15$ mol trona, $y=0.05$ mol $\ce{NaOH}$ .

Assumes the $\ce{HCO3-}$ from trona is unaffected by the added $\ce{NaOH}$ (ignoring the acid–base reaction between $\ce{OH-}$ and $\ce{HCO3-}$ ), so sets trona $x=0.10$ mol directly to match the target $\ce{HCO3-}$ . Then uses $3x+y=0.50$ to get $y=0.20$ mol $\ce{NaOH}$ — this overcounts $\ce{HCO3-}$ because it forgets that some of it is consumed by the base.

Reverses the direction of the acid–base reaction, treating added $\ce{NaOH}$ as if it were regenerating $\ce{HCO3-}$ from $\ce{CO3^2-}$ instead of consuming it, giving the flipped balance $x-y=-0.10$ (paired with $3x+y=0.50$ ). This sign error yields $x=0.05$ mol trona and $y=0.35$ mol $\ce{NaOH}$ , the opposite of the correct amounts.

Incorrectly assumes the $\ce{HCO3-}$ : $\ce{CO3^2-}$ ratio delivered by trona is fixed and cannot be altered, missing that adding a strong base like $\ce{NaOH}$ shifts some $\ce{HCO3-}$ into $\ce{CO3^2-}$ , which actually allows the desired concentrations to be reached with the two reagents.

Concept · Acid-base speciation and mass balance in mixed carbonate/bicarbonate solutions