Question

Lets do a little more practice with titration calculations: If it takes 83 mL of a 0.45 M standard solution to neutralize 235 mL of an analyte solution, then the concentration of the analyte solution is Type your answer here M. It takes 12.5 mL of a 0.30 M HCl solution to neutralize 285 mL of NaOH solution, then the concentration of the NaOH solution is Type your answer here M.

Answer 1

If it takes 83 mL of a 0.45 M standard solution to neutralize 235 mL of an analyte solution, then the concentration of the analyte solution is 0.16 M

Step-by-step explanation:

for standard solution(NaOH):

0.45 M = mol/litre

therefore, n(NaOH) = 0.45mole x 83 / 1000

n(NaOH) = 0.037 moles of NaOH.

1NaOH +1HCl ⇒ 1NaCl + 1H2O

1 mole of NaOH = 1 mole of HCl

thus, n(HCl) = 0.037 moles

concentration of HCl = mole/litre

                                  = 0.037 / 0.235

                                  = 0.157

                                  ≈ 0.16 M

It takes 12.5 mL of a 0.30 M HCl solution to neutralize 285 mL of NaOH solution, then the concentration of the NaOH solution is  0.0131 M.

Step-by-step explanation:

0.013 M NaOH

HCL + NaOH = NaCL + water

The mole ratio of acid and base is 1:1

Molarity is equal to the moles of solute divided by the liters of solution

Solute mole = 12.5 × 10^-3 × 0.3 = 0.003075

To neutralize 0.00375mol of HCl, 0.00375 mol of NaOH will be required. Since molarity is moles per liter, divide this amount of moles by the volume in liters:

Molarity = 0.00375/285 × 10^-3

Molarity = 0.0131 M

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