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1.How many mL of 0.500 M HCl should be added to 155 mL of 0.0500 M...

1.How many mL of 0.500 M HCl should be added to 155 mL of 0.0500 M benzylamine to form a buffer with pH = 8.52? C7H7NH2 pKb = 4.35 C7H7NH2 (aq) + H+ (aq)  C7H7NH3 + (aq) This structure of benzylamine has a carbon at the junction of each line with 1 hydrogen on each 3-bonded ring C and 2 hydrogens on the C in the side chain attached to the ring carbon. Note that the N atom has 3 bonds and a lone pair.

2.Write the formula for the conjugate acid of benzylamine.

Solutions

Expert Solution

benzylamine millimoles = 155 x 0.05 = 7.75

millimoles of HCl = 0.5 x V = 0.5 V

C7H7NH2 (aq) + H+ (aq) -----------------> C7H7NH3 + (aq)

     7.75                0.5 V                                 0           

    7.75 - 0.5V         0                                     0.5V

pH = 14 – { pKb + log[salt/ base ]}

8.52 = 14 - {4.35 + log [0.5V / 7.75 - 0.5V]

[0.5V / 7.75 - 0.5V] = 13.49

V = 14.43 mL

volume of HCl added = 14.43 mL

queen_honey_blossom answered 2 years ago
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