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QUESTION 4 How many millimoles (mmol) of H 30 * are in the acid solution at the beginning of the titration before any base is


QUESTION 3 To answer Questions through 7. consider a situation in which 25.00 mL of HCl (a strong acid) is titrated with NaOH
QUESTION 7 Use your answers to Questions 4 and 6 to calculate the molarity of the NaOH solution used in this titration. Type
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Answer #1

3-

The neutralization reaction between HCl and NaOH is-

HCl + NaOH ---------> NaCl + H2O

that means 1 mole of HCl will neutralize 1 mole of NaOH.

Now given before start of titration,

pH of the solution = 1.02

Again we know

pH = -log [H3O+]

So

[H3O+] = 10-pH  

= 10-1.02  

=  0.095 M

= 0.095 mole/L

= 9.5 * 10-2 mole/L

4-

Now given volume of HCl taken = 25 mL

So we can say volume of H3O+ present = 25 mL  

Because HCl is a strong acid and it will dissociate completely to H3O+ . So what ever the values for HCl can be exactly taken for that of H3O+

Again calauclated concentration of H3O+ = 9.5 * 10-2 mole/L

So moles of H3O+ present = concentration * volume

= 9.5 * 10-2 mole/L * 25 mL  

=  9.5 * 10-2 mole/ 1000 mL * 25 mL  

= 0.002375‬ moles

So milimoles of H3O+ present = 0.002375‬ moles * 1000

= 2.375‬ milimoles

For further calculations, no titration curve is provided.

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