2. Use your assigned set of data to determine the molarity of HCL Record the res
ID: 591620 • Letter: 2
Question
2. Use your assigned set of data to determine the molarity of HCL Record the result to two decimal places a. 40.0 mL of 0.10 M NaOH requires 30.0 mL of HCI for neutralization. b. 30.0 mL of 0.20 M NaOH requires 20.0 mL of HCI for neutralization. c. 24.0 mL of 0.40 M NaOH requires 36.0 ml of HCI for neutralization. d. 18.0 mL of 0.30 M NaOH requires 27.0 mL of HCI for neutralization. e. 32.0 mL of 0.50 M NaOH requires 16.0 mL of HCI for neutralization. f. 23.0 mL of 0.40 M NaOH requires 29.0 mL of HCI for neutralization. g. 6.0 mL of 0.60 M NaOH requires 34.0 mL of HCI for neutralization. h. 35.0 mL of 0.20 M NaOH requires 14.0 mL of HCI for neutralization . Suppose you suddenly realize you did not add the phenolphthalein to the acid solution while you are in to the mixture, the solution: a. remains colorless b. becomes pink. Should you start over or can you rescue the current titration? Explain your answer.Explanation / Answer
2)
a) 40 mL of 0.1 M NaOH ;
No. moles of NaOH = molarity x volume in L
= 0.1 mol/L x 0.040 L
= 0.004 mol
Therefore, no. of moles of HCl require is = 0.004 mol
molarity of HCl solution = no. of moles/volume in L
= 0.004 mol / 0.030 L
= 0.133 M
Therefore, molarity of HCl = 0.133 M
b) 30 mL of 0.2 M NaOH ;
No. moles of NaOH = molarity x volume in L
= 0.2 mol/L x 0.03 L
= 0.006 mol
Therefore, no. of moles of HCl require is = 0.006 mol
molarity of HCl solution = no. of moles/volume in L
= 0.006 mol / 0.020 L
= 0.3 M
Therefore, molarity of HCl = 0.3 M
c) 24 mL of 0.4 M NaOH ;
No. moles of NaOH = molarity x volume in L
= 0.4 mol/L x 0.024 L
= 0.0096 mol
Therefore, no. of moles of HCl require is = 0.0096 mol
molarity of HCl solution = no. of moles/volume in L
= 0.0096 mol / 0.036 L
= 0.267 M
Therefore, molarity of HCl = 0.267 M
d) 18 mL of 0.3 M NaOH ;
No. moles of NaOH = molarity x volume in L
= 0.3 mol/L x 0.018 L
= 0.0054 mol
Therefore, no. of moles of HCl require is = 0.0054 mol
molarity of HCl solution = no. of moles/volume in L
= 0.0054 mol / 0.027 L
= 0.2 M
Therefore, molarity of HCl = 0.2 M
e) 32 mL of 0.5 M NaOH ;
No. moles of NaOH = molarity x volume in L
= 0.5 mol/L x 0.032 L
= 0.016 mol
Therefore, no. of moles of HCl require is = 0.016 mol
molarity of HCl solution = no. of moles/volume in L
= 0.016 mol / 0.016 L
= 1 M
Therefore, molarity of HCl = 1.0 M
f) 23 mL of 0.4 M NaOH ;
No. moles of NaOH = molarity x volume in L
= 0.4 mol/L x 0.023 L
= 0.0092 mol
Therefore, no. of moles of HCl require is = 0.0092 mol
molarity of HCl solution = no. of moles/volume in L
= 0.0092 mol / 0.029 L
= 0.317 M
Therefore, molarity of HCl = 0.317 M
g)6 mL of 0.6 M NaOH ;
No. moles of NaOH = molarity x volume in L
= 0.6 mol/L x 0.006 L
= 0.0036mol
Therefore, no. of moles of HCl require is = 0.0036 mol
molarity of HCl solution = no. of moles/volume in L
= 0.0036 mol / 0.034 L
= 0.106 M
Therefore, molarity of HCl = 0.106 M
h) 35 mL of 0.2 M NaOH ;
No. moles of NaOH = molarity x volume in L
= 0.2 mol/L x 0.035 L
= 0.007 mol
Therefore, no. of moles of HCl require is = 0.007 mol
molarity of HCl solution = no. of moles/volume in L
= 0.007 mol / 0.014 L
= 0.5 M
Therefore, molarity of HCl = 0.5 M
The phenolphthalein indicator is colorless in acidic solution, whereas in the basic medium the solution turns to pink color.
Therefore, in the case of
a) the solution remains colorless;
means, that the endpoint is not yet reached. So, there is no problem in continuing the titration in this case.
b) becomes pink;
In this case, the endpoint is already reached. So, there is no point in continuing the titration.
Hence, you have to start fresh solution again in this case.
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