What is the pH of a solution in which 35 mL Of 0.10 M NaOH is added to 25 mL of

Question

What is the pH of a solution in which 35 mL Of 0.10 M NaOH is added to 25 mL of 0.10 M HCI? A 1.24-g sample of benzoic acid was dissolved in water to give 50.0 mL of solution. This solution was titrated with 0.180 M NaOH. What was the pH of the solution when the equivalence point was reached? A 0.400-g sample of propionic acid was dissolved in water to give 50.0 mL of solution. This solution was titrated with 0.150 M NaOH. What was the pH of the solution when the equivalence point was reached? Find the pH of the solution obtained when 32 mL of 0.087 M ethylamine is titrated to the equivalence point with 0.15 M HCI? What is the pH at the equivalence point when 22 mL of 0.20 M hydroxylamine is titrated with 0.10 M HCI? A 50.0-mL sample of a 0.100 M solution of NaCN is titrated by 0.200 M HCI. Kb for CN^- is 2.0 times10^-5.Calculate the pH of the solution: a prior to the start of the titration; b after the addition of 15.0 mL of 0.200 M HCI; c at the equivalence point; d after the addition of 30.00 mL of 0.200M HCI. Sodium benzoate, NaC_7H_5O_2, is used as a preservative in foods. Consider a 50.0-mL sample of 0.250 M pH of the solation: a when no HBr has been added; b after the addition of 50.0mL of the HBr solution; c at the equivalence point; d after the addition of 75.00 mL of the HBr solution. The K_b value for the benzoate ion is 1.6times10^10. Calculate the pH of a solution obtained by mixing 25.00 mL of 0.19 M NH_3 with 25.00 mL of 0.060 M HCI.16.94 Calculate the pH of a solution obtained by mixing 35.0mL of 0.15 M acetic with 27.0 mL of 0.10 M sodium acetate.

Explanation / Answer

Ans 16.89

The salt formed during the titration will be , C2H5NH3+Cl-

First determine the molarity of salt formed ,

M = 0.087 / 0.032 = 0.002784 mol

now determine the volume of acid, using the number of moles of salt and molarity

V = 0.02784 / 0.15 = 18.56 ml

Now the volume of solution after titration , during equivalece point will be ,

32 ml + 18.56 ml = 50.56 ml

Now determine the molarity of resulting solution ,

M = 0.002784 / 0.05056 = 0.05506 M

The salt undergoes hydrolysis,

C2H5NH3+ + H2O ---> C2H5NH2 + H3O+

So this resulting soluition you see is acidic

ka of C2H5NH3+ = 1.56 x 10-11

ka = ([C2H5NH2] [H3O+] / [C2H5NH3+])

1.56 x 10-11 = ( x. x ) / ( 0.05506 - x)

x <<< 0.05506 so ignore it

x = square root of [( 1.56 x 10-11) ( 0.05506)]

x = 9.268 x 10-7

pH = - log [H3O+]

pH = -log 9.268 x 10-7

pH = 6.03

So the ph of the solution at equivalence point is 6.03 ( slightly acidic )

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