Chemistry is not difficult

Question

Help me with this exercise!!!

Mix 50 ml of 0.1M HCl solution with 100 ml of 0.2M NaOH solution. Calculate the molar concentration of the ions in the solution after the reaction.

Drift Hunters

Answer 1

Firstly, When you mix HCl (a strong acid) with NaOH (a strong base), they undergo a neutralization reaction:  

HCl(aq) + NaOH(aq) → NaCl(aq) + H₂O(l)

. Moles of Reactants:

HCl: 0.1 M * 0.050 L = 0.005 moles

NaOH: 0.2 M * 0.100 L = 0.020 moles

2. Limiting Reactant: HCl (all consumed)

3. Moles After Reaction:

NaCl formed: 0.005 moles

NaOH remaining: 0.015 moles

4. Total Volume: 50 ml + 100 ml = 150 ml = 0.150 L

5. Ion Concentrations:

[Na⁺] = (0.005 mol + 0.015 mol) / 0.150 L = 0.133 M

[Cl⁻] = 0.005 mol / 0.150 L = 0.033 M

[OH⁻] = 0.015 mol / 0.150 L = 0.100 M

Answer 2

Look at the reaction taking place. It is a neutralization reaction between a strong acid (HCl) and a strong base (NaOH).

HCl(aq) + NaOH(aq) <==> NaCl(aq) + H₂O(l) .. balanced equation

Now, we will find the moles of each reactant present:

moles HCl = 50 ml x 1 L / 1000 ml x 0.1 mol / L = 0.005 mols HCl

moles NaOH = 100 ml x 1 L / 1000 ml x 0.2 mol / L = 0.02 mols NaOH

Clearly the NaOH is in excess, meaning that all of the HCl will react (leaving no HCl left over), and there will be excess NaOH after the reaction is complete. We can view this easily using an ICE table...

HCl(aq) + NaOH(aq) <==> NaCl(aq) + H₂O(l)

0.005.........0.02........................0.................0............Initial

-0.005......-0.005....................+0.005........................Change

0...............0.015......................0.005........................Equilibrium

The final volume will be 50 ml + 100 ml = 150 mls = 0.150 L

The final concentrations will be...

[HCl] = 0

[NaOH] = 0.015 mol / 0.150 L = 0.10 M

[NaCl] = 0.005 mol / 0.150 L = 0.033 M

So, the final concentrations of the IONS will be...

[Na⁺] = 0.10 + 0.033 = 0.13 M

[Cl^-] = 0.033 M

[OH^-] = 0.10 M

NOTE: these calculations do NOT include the [H⁺] and [OH^-] contributed by the autoionization of the H₂O as they would be insignificant. Also, they are not corrected for significant figures, which should be only 1 sig.fig. based on the given molarities.