What is the hydroxide-ion concentration in a solution formed by combining 200.mL of 0.15 M HCl with 300.mL of 0.091 M NaOH at 25°C?
HCl(aq) + NaOH(aq) $\to$ NaCl(aq) + H₂O(l)

Question

What is the hydroxide-ion concentration in a solution formed by combining 200.mL of 0.15 M HCl with 300.mL of 0.091 M NaOH at 25°C? HCl(aq)+ NaOH(aq) $\to$ NaCl(aq)+ H₂O(l)

Quizplus · Accepted Answer

First, find the amount of moles of HCl and NaOH:
moles HCl = 0.15 M x 0.2 L = 0.03 moles
moles NaOH = 0.091 M x 0.3 L = 0.0273 moles

Since HCl and NaOH react in a 1:1 ratio, all of the NaOH will react with an equal amount of HCl. This leaves 0.003 moles of HCl unreacted. Therefore, the amount of moles of OH- produced is equal to the amount of moles of NaOH added:
moles OH- = 0.0273 moles

Total volume of solution = 200 mL + 300 mL = 0.5 L

Now use the definition of concentration and the amount of moles of OH- produced to find the hydroxide-ion concentration:
[OH-] = moles OH- / total volume of solution
[OH-] = 0.0273 moles / 0.5 L = 0.0546 M = 1.9×10^-12 M (scientific notation)

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