Question

1. A 100.0-mL sample of 0.500 M sodium hydroxide is titrated with 0.100 M nitric acid....

1. A 100.0-mL sample of 0.500 M sodium hydroxide is titrated with 0.100 M nitric acid. Calculate the pH before the titration begins.

2. A 1.0-L buffer solution contains 0.100 mol HCN and 0.100 mol LiCN. The value of Ka for HCN is 4.9 x 10-10. Because the initial amounts of acid and conjugate base are equal, the pH of the buffer is equal to pKa = -log (4.9 x 10-10) = 9.31. Calculate the new pH after the addition of 0.010 mol of HNO3 to the buffer. Ignore any change of volume brought about by this addition.

3. A 100.0-mL sample of 0.100 M HCHO2 is titrated with 0.200 M NaOH. Calculate the pH after adding 30.00 mL of the base.

1.

Before the titration begins, there is only NaOH present in the solution.

NaOH is a strong base. Hence, it completely ionizes in water.

Thus, 0.500 M of NaOH ionizes to give 0.500 M of OH- in aqueous solution.

Thus,

pOH = - log[OH-]

= - log(0.500)

= 0.301

Now,

pH + pOH = 14

or, pH = 14 - pOH

= 14 - 0.301

= 13.7

Therefore, the pH before the titration begins = 13.7

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