For each of the following solutions, calculate the initial pH and the final pH a

Question

For each of the following solutions, calculate the initial pH and the final pH after adding 0.010 mol of NaOH.

A.

For 260.0mL of pure water, calculate the initial pHand the final pH after adding 0.010 mol of NaOH.

B.

For 260.0mL of a buffer solution that is 0.240M in HCHO2 and 0.305M in KCHO2, calculate the initial pH and the final pH after adding 0.010 mol of NaOH.

C.

For 260.0mL of a buffer solution that is 0.325M in CH3CH2NH2 and 0.300M in CH3CH2NH3Cl, calculate the initial pH and the final pH after adding 0.010 mol of NaOH.

Explanation / Answer

Answer : Here the concept of buffer is used .

a} Before addition of NaOh

as we know that Kw [ for water ] = 10-14= [H+] [OH-]

and concentration of OH- and H+ is same in pure water hence concentration of H+ ion is 10-7

Now , the Ph = -log[H+] = - log[10-7] = 7

Now we know that concentration of H+ = 10-7

So the number of moles in 260 ml = 10-7 * 260 /1000 = 2.6 * 10-8 mol

H2O <---------------------------------> H + + Oh -

initially 2.6 * 10-8    2.6 * 10-8

Now after addition of NaoH the number of moles of OH- becomes 2.6 * 10-8 + 0.010 mol

hence total moles of OH- = 0.01000003

Now , toatl volume is = 260 ml

Hence concentration of OH- = 0.01000003 * 1000 / 260 = 0.03846

now PoH = -log[oH-] = 1.4149

hence PH = 14 - POH = 12.5851

Hence the final PH = 12.5851 .

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