art A hat mass of HCl can this buffer neutralize before the pH falls below 9 00
ID: 636392 • Letter: A
Question
art A hat mass of HCl can this buffer neutralize before the pH falls below 9 00 xpress the mass in grams to three significant figures 1.82.10-3 Submit Incorrect; Try Again To solve for the mass of HCl required to bring the pH to 9 00. set up a Henderson-Hasselbalch oquation (pH increase the value of the denominator and decrease the value of the numerator in the log fraction. pK, + log basel/acid]) Adding acid will The value of K is given, so you willl need to convert the value either to pK, then pK, or to K, then pK Part B f the same volume of the buffer were 0250 Min NHs and 0 390 Min NH, Br what mass of HCl could be handied before the pH tls below 9 00? Express the mass in grams to three significant figures. m 0.0181 SubmitExplanation / Answer
Part A
Basic buffer system
pOH of basic buffer = pkb + log((NH4Br+HCl)/(NH-HCl))
pOH = 14-pH
= 14-9
= 5
pkb OF nh3 = -log(1.76*10^-5) = 4.75
No of mol of NH4Br = M*V = 0.135*0.13 = 0.01755 mol
No of mol of NH3 = 0.1*0.13 = 0.013 mol
no of mol of HCl can be added = x mol
5 = 4.75+log((0.01755+x)/(0.013-x))
x = 0.002
No of mol of HCl can be neutralized = 0.002 mol
amount of HCl can be neutralized = n*M
= 0.002*36.5 = 0.073 g
part B
Basic buffer system
pOH of basic buffer = pkb + log((NH4Br+HCl)/(NH-HCl))
pOH = 14-pH
= 14-9
= 5
pkb OF nh3 = -log(1.76*10^-5) = 4.75
No of mol of NH4Br = M*V = 0.39*0.13 = 0.0507 mol
No of mol of NH3 = 0.25*0.13 = 0.0325 mol
no of mol of HCl can be added = x mol
5 = 4.75+log((0.0507+x)/(0.0325-x))
x = 0.00255
No of mol of HCl can be neutralized = 0.00255 mol
amount of HCl can be neutralized = n*M
= 0.00255*36.5 = 0.093 g
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