Question

# A 130.0 −mL buffer solution is 0.110 M in NH3 and 0.130 M in NH4Br. If...

A 130.0 −mL buffer solution is 0.110 M in NH3 and 0.130 M in NH4Br.

If the same volume of the buffer were 0.265 M in NH3 and 0.390 M in NH4Br, what mass of HCl could be handled before the pH fell below 9.00?

pH = 9, so pOH = 14-9 = 5

this means

intial pOH, from henderson hasselbach equation;

pOH = pKb + log(NH4+/NH3)

pOH initial = 4.75 + log(0.13/0.11)

pOH initial = 4.8225

we need to drop below 9, i.e. above of 5

so

mass of Hcl --> mmol of HCl

when we add HCl we increase NH4+ since we neutralize NH3 base

NH4+ + H+ = 0.13*130 + x

NH3 + OH+ = 0.11*130- x

pOH required= 4.75 + log((16.9+ x)/( 14.3- x))

5 = 4.75 + log((16.9+ x)/( 14.3- x))

(16.9+ x)/( 14.3- x) = 10^(5-4.75) = 1.7782

1.7782(14.3-x) = 16.9+x

1.7782(14.3) - 1.7782x = 16.9 + x

1.7782(14.3) - 16.9 = (1+1.7782)*x

8.52826 = 2.7782x

x = 8.52826/2.7782 = 3.0697 mmol

mass = mmol*MW = 3.0697*36.5 = 112.0440 mg of HCl required or 0.112 g of HCl

#### Earn Coins

Coins can be redeemed for fabulous gifts.
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}