Calculate the pH of the solution that results when 32.0 mL of 0.1670 M NH_3 is (

Question

Calculate the pH of the solution that results when 32.0 mL of 0.1670 M NH_3 is (This problem requires values in your textbook's specific appendices, which you can access through the owLv2 MindTap Reader. You should not use the OWLv2 References' Tables to answer this question as the values will not match.) diluted to 95.0 mL with distilled water. pH = mixed with 16.0 mL of 0.334 M HCl solution. pH = mixed with 16.0 mL of 0.400 M HCl solution. pH = mixed with 16.0 mL of 0.400 M NH_4 Cl solution. pH = mixed with 16.0 mL of 0.320 M HCl solution. pH =

Explanation / Answer

a)

millimoles of NH3 = 32 x 0.167 = 5.344

volume of distilled water = 95.00 mL

M1 V1 = M2 V2

0.167 x 32 = M2 x 95

M2 = 0.05625 M

concentration of NH3 = 0.05625 M

pKb = 4.74

pOH = 1/2 (pKb - log C)

        = 1/2 (4.74 - log 0.05625)

       = 3.00

pH = 11.00

b)

millimoles of HCl = 16 x 0.334 = 5.344

millimoles of NH3 = 32 x 0.167 = 5.344

this is equivalence point :

here salt only remians.

salt concentration = 5.344 / (32 + 16) = 0.1113 M

pH = 7 - 1/2 (pKb + log C)

    = 7 - 1/2 (4.74 + log 0.1113)

   = 5.11

pH = 5.11

c)

millimoles of HCl = 16 x 0.4 = 6.4

here strong acid remains = 6.4 - 5.344 = 1.056

[H+] = 1.056 / 32 + 16 = 0.022 M

pH = -log 0.022

pH = 1.66

d)

millimoles of NH4Cl = 16 x 0.4 = 6.4

pOH = pKb + log [salt / base]

        = 4.74 + log [6.4 / 5.344]

        = 4.82

pH = 9.18

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