I am asked to find the concentration of the unknown HCl solution for each trial

Question

I am asked to find the concentration of the unknown HCl solution for each trial using my calculated avg concentration of titrant used (NaOH .1002 M). I have no idea how to begin. Please help!!!! Itration of unknown solution Unknown ID: Balanced equation for neutralization on: Rea 4 Trial 1 Tial 2 Trial 3 Initial buret reading: aluding11.2 m 123n 11 2lo ml Final buret reading: Volume of titrant used: Volume of unknown used: Concentration of unknown: Average concentration of unknown: Deviation: Relative average deviation: Clearly show all calculations for the titration of the unknown acid: 21 mL IL23mL ll:210 mL 0

Explanation / Answer

The concentration of unknown can be known from a concentration and volume of known solution by the following formula, M1V1/n1=M2V2/n2

WHere, M1and V1 are molarity and volumes of solution A and n1 is no. of moles of A

M2and V2 are molarity and volumes of solution B and n2 is no. of moles of B

Chemical equation is HCl + NaOH ----> NaCl + HCl

here 1mole HCl = 1mole NaOH (n1 & n2 =1)

Therefore, equation may be simplified as M1V1=M2V2

Let, M1=Molarity of NaOH = 0.1002M, V1 = Volume of NaOH consumed = 11.21mL

M2 = Molarity of HCl =?, V2= Volume of HCl taken = 25.0mL

Now, 0.1002M*11.21mL = M2*25.0mL

Trail1: M2=0.1002M*11.21mL/25.0mL = 0.0449M

Trail:2 Trail1: M2=0.1002M*11.23mL/25.0mL = 0.045M

Trail3: M2=0.1002M*11.21mL/25.0mL = 0.0450M

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