Hello, I am wondering that could you please check my solutions to the problems?

Question

Hello, I am wondering that could you please check my solutions to the problems?

Cf. This experiment is "Strong Acid + Strong Base" titration. 5.00 mL of the M(OH)2 is given, and concentration of HCl from the bottle is 0.0512 M HCl and the calculated concentration of M(OH)2 is 0.0720 M M(OH)2.

Could you compare your answers with my solutions above and tell me what are the incorrect?

Report Question 1: Using the concentration values determined in this investigation for M(OH)2 and HCl calculate the pH for a strong acid + strong base titration in which 5.00 mL of the M(OH).2 was transferred via pipet to a beaker and HCl was added from the buret. Calculate the pH: a) before any HCl is added b) after the addition of 4.00 mL HC c) after the addition of 9.00 mL HCl d) 4.00 mL beyond the equivalence point Show your work, as applicable.

Explanation / Answer

M(OH)2:

Volume = 5 mL = 0.005 L
Molarity = 0.0720 M

Moles = Molarity x Liter
So, moles of M(OH)2 = 0.0720 M x 0.005 L = 0.144 moles

Now, M(OH)2 will dissociate completely into
M(OH)2   <----->   M2+   +   2OH-

So, [OH-] = 2 x 0.0720 M = 0.114 M

pOH = - log [OH-]
         = - log (0.114)
         = 0.943

pH = 14 – pOH
      = 14 - 0.943
      = 13.057

(b)

Moles = Molarity x Liter

M(OH)2: Volume = 5 mL = 0.005 L, Molarity = 0.0720 M
So, moles of M(OH)2 = 0.0720 M x 0.005 L = 0.00036 moles

HCl: Volume = 4 mL = 0.004 L, Molarity = 0.0512 M
So, moles of HCl = 0.0512 M x 0.004 L = 0.0002048moles

Now,

The reaction equation is expressed as

2 HCl +   M(OH)2   ------>    MCl2   +   2 H2O

So, 2 moles of HCl reacts with 1 mole of M(OH)2
or, 1 mole of HCl reacts with 1/2 moles of M(OH)2
or, 0.0002048 moles of HCl reacts with (0.0002048/2) moles of M(OH)2
or, 0.0002048 moles of HCl reacts with 0.0001024 moles of M(OH)2

So, moles of M(OH)2 left = 0.00036 moles – 0.0001024 moles
                                          = 0.0002576 moles

Total volume = 5 mL + 4 mL = 9 mL = 0.009 L

So, [M(OH)2] = 0.0002576 moles / 0.009 L= 0.0286 M

Hence, [OH-] = 2 x 0.0286 M = 0.0572 M

pOH = - log [OH-]
         = - log (0.0572)
         = 1.24

pH = 14 – pOH
      = 14 - 1.24
      = 12.76

(d)

4 mL beyond the equivalence point:

At equivalence point, moles of M(OH)2 = moles of HCl

Moles of M(OH)2 = 0.00036 moles = moles of HCl

Molarity of HCl = 0.0512 M

So, volume of HCl required to reach equivalence point = Moles / Molarity
                                                                                        = 0.00036 moles / 0.0512 M
                                                                                        = 0.00703 L
                                                                                        = 7.03 mL

Now, 4 mL of HCl of 0.0512 M is added beyond equivalent point. At equivalent point all the strong base, M(OH)2, has neutralized. And hence, only left with HCl

So, moles of HCl = 0.0512 M x 0.004 L = 0.0002048 moles

Total volume = 5 mL + 7.03 mL + 4 mL = 16.03 mL = 0.01603 L

So, [HCl] = 0.0002048 moles / 0.01603 L = 0.0128 M

pH = - log [H+]
      = - log (0.0128)
         = 1.89

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