Question

Amy prepared her base solution by adding 16 mL of 6.1 M NaOH to her 500 mL bottle and then filling it to the shoulder. She then standardized her base solution by titrating it with a 0.216 M HCl solution. She filled one buret with her NaOH solution and another buret with the HCl solution. The initial readings on the burets were 1.74 mL (NaOH) and 1.43 mL (HCl). She then dispensed some HCl into an Erlenmeyer flask, added a couple drops of indicator and then added NaOH until the endpoint was reached. At that point, the readings on the burets were 26.41 mL (NaOH) and 26.60 mL (HCl). What is the concentration (in moles per liter) of the NaOH solution that Amy made? Type your answer rounded to the 4th decimal place (N.NNNN) without units. Note: molarity (M) is simply an abbreviation for moles per liter. For example, a 3.4 M NaBr solution would have 3.4 moles of NaBr per liter of solution - molarity is just a conversion factor to go between moles and volume.

Answer 1

Reaction between NaOH and HCl is written as

NaOH + HCl ----------> NaCl + H2O

$\frac{M_{1}V_{1}}{n_{1}} = \frac{M_{2}V_{2}}{n_{2}}$

Let

M₁​​​ = Molarity of NaOH = ?

V₁ = Volume of NaOH = Final Burette volume - initial volume of burrtte = 26.41- 1.74 = 24.67 mL

n_{​​​​​​1} = number of moles of NaOH = 1

M_{​​​​​​2} = Molarity of HCl = 0.216 M

V_{​​​​​​2}= Volume of HCl = Final Burette volume - initial volume of burrtte = 26.60- 1.43 = 25.17 mL

n_{​​​​​2} = number of moles of HCl= 1

$M_{1} = \frac{M_{2}V_{2}}{n_{2}}*\frac{n_{1}}{V_{1}}$

$M_{1} = \frac{0.216 M*25.17 mL}{1 mole}*\frac{1 mole}{24.67 mL}$

M_{​​​​​​1} = 0.2204 M

After standardisation, concentration of NaOH = 0.2204 M

Dilution formule is Mv, = M₂ V₂ M : Initial molarity of NaOH = 6.1M Vilnitial volume of NaoH - 16 mL. M. Final Molarity of Naoth = ? - Final volume of Noot-500m M, V 16 mt x6Н M₂ - - Soome 0. 1952 M The Concent No Reaction between NaOH & He is NaOH + Hd - Nadt H₂O Volume MIVI - M2 V₂ n M2 M, : Molarity of Noolt ? Ve volume of NaOH = final Burette volumen Inition = 26.41-1.74 = 24.67 mL Noot Je M = Molasity of Hel=0.216M. V: Volume of Hel consurned - final volume - nehal volum = 26.60-1.43 =25.17 = No of moles of el= 1 colt i mole of = 1 1: Number of me

M2V & Mi- me ty 0.216 M X 25.17mL I mole 24.67 0.2204 M Actually its loncentration of NaOH. prepared should be 0.1952 m. After standerdising, the concentration of NaOH is 0.2204M.