Question

The following data were collected at the endpoint of a titration performed to find the molarity of an HCl solution.Volume of acid (HCl) used = 26.8 mLVolume of base (KOH) used = 19.5 mLMolarity of standard base (KOH) = 0.70 MWhat is the molarity of the acid solution?Select one:a.0.51 Mb.0.74 Mc.1.2 Md.1 M

Answer 1

The reaction is

`\text{ KOH + HCl }\rightarrow\text{ KCl + H}_2\text{O}`

moles of acid

`\text{ 19.5 mL KOH }*\frac{1\text{ L}}{\text{ 1000 mL}}*\frac{\text{ 0.7 mol KOH}}{1\text{ L KOH}}*\frac{1\text{ mol HCl}}{1\text{ mol KOH}}=\text{ 0.01365 mol HCl}`

Molarity of acid

`\text{ M = }\frac{\text{ moles of acid}}{\text{ Volume of acid}}=\text{ }\frac{\text{ 0.01365 mol}}{\text{ 0.0268 L}}=\text{ 0.51 M}`

The answer is: the molarity of the acid solution is 0.51 M