Question
In: Chemistry
Consider a buffer made by adding 132.8 g of NaC₇H₅O₂ to 300.0 mL of 2.15 M...
Consider a buffer made by adding 132.8 g of NaC₇H₅O₂ to 300.0 mL of 2.15 M HC₇H₅O₂ (Ka = 6.3 x 10⁻⁵) What is the pH of the buffer after 0.750 mol of H⁺ have been added?
Solutions
Expert Solution
no of moles of NaC7H5O2 = W/G.M.Wt
= 132.8/144 = 0.923moles
no of moles of HC7H5O2 = molarity * volume in L
= 2.15*0.3 = 0.645moles
PKa = -logKa
= -log(6.3*10^-5)
= 4.2
no of moles of NaC7H5O2 after adding 0.750 mol of H^+ = 0.923-0.75 = 0.173 moles
no of moles of HC7H5O2 after adding 0.750mol of H^+ = 0.645+0.75 = 1.395moles
PH = Pka + log[NaC7H5O2]/[HC7H5O2]
= 4.2 + log0.173/1.395
= 4.2 -0.9065
= 3.2935>>>>answer
queen_honey_blossom answered 1 year agoRelated Solutions
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