**Please show all work and equations of how the solution wasreached in detail...

Question

**Please show all work and equations of how the solution wasreached in detail...Thank you**

2) In an acid-base titration, 22.13 mL of an NaOH solutionare needed to neutralize 24.65 mL of a 0.1094 M HCl solution. To find the molarity of the NaOH solution, we can use the followingprocedure:

a. First not the value of MH+ in the HClsolution.
                                                                                                            ______________M

b. Find MOH- in the NaOHsolution.            
                                                                                                           _______________M

c. Obtain M NaOH from MOH- .
                                                                                                            _______________M

Explanation / Answer

(a)    Molarity of H+ in the HClsolution MH+ = molarity of HCl   (since1 mole of HCl gives 1 mole of H+ ions)                                                             = 0.1094 M . (b) Molarity of H+, M1 = 0.1094 M volume, V1 = 24.65 mL Molarity of OH- = M2 Volume, V2 = 22.13 mL M1 V1 = M2 V2 Molarity of OH- , M2 = M1 V1 / V2                                  = 0.1094 M * 24.65 mL / 22.13 mL                                   =0.122 M . (c)   One mole of NaOH gives one mole of OH-moles. Hence MNaOH = MOH- = 0.122 M

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