You need to prepare 100.0 mL of a pH=4.00 buffer solution using 0.100 M benzoic

Question

You need to prepare 100.0 mL of a pH=4.00 buffer solution using 0.100 M benzoic acid (pKa = 4.20) and 0.220 M sodium benzoate. How much of each solution should be mixed to prepare this buffer? 500.0 mL of 0.160 M NaOH is added to 535 ml of 0.250 M weak acid (K, = 4.49 * 10-6). What is the pH of the resulting buffer? Silver chromate is sparingly soluble in aqueous solutions. The Ksp of Ag2CrC>4 What is the solubility (in mol/L) of silver chromate in 1.30 M potassium chromate aqueous solution? in 1.30 M silver nitrate aqueous solution? in pure water?

Explanation / Answer

pH = pka + log [sod benzoate]/[benzoic acid]

4 = 4.2 + log [ sod benzoate]/[benzoic acid]

[sod benzoate] = 0.631 [benzoic acid]

sod benzoate moles = 0.631 benzoic acid moles

0.22 x V = 0.631 ( 0.1 x ( 0.1-V) )

0.22V = 0.00631-0.0631V

V = 0.0223 , hence sodium benzoate vol = 0.0223 x 1000 ml = 22.3 ml , benzoic acid vol = 100-22.3 = 77.7 ml

2) pka = -log ( 4.49 x 10^-5) = 4.35,

weak acid moles after reacting with NaOH = ( 0.535 x 0.25) - ( 0.5 x 0.16) = 0.05375

A- moles = ( 0.5x0.16) = 0.08 , total vol = 0.535+0.5 = 1.35 liters

pH = 4.35 + log ( 0.08/1.35)/( 0.05375/1.35)   

= 4.52

3) Ksp = [Ag2+]^2 [CrO42-] = (2s)^2 x s = 4s^3

in presence of 1.3 M K2CrO4 we have [CrO4] = 1.3 M

hence Ksp = 1.12 x 10^-12 = (2s)^2 x ( s+ 1.3)

solubility s = 4.64 x 10^-7 M

(b) 1.3 AgNO3 we have [Ag+] = 1.3 M

hence 1.12 x 10^-12 = ( 2s+ 1.3) ^2 x s

s = 6.63 x 10^-13 M

(c) in water 1.12 x 10^-12 = 4s^3

s = 6.54 x 10^-5 M

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