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A 130.0 mL buffer solution is 0.105 molL−1 in NH3 and 0.125 molL−1 in NH4Br. What...

A 130.0 mL buffer solution is 0.105 molL−1 in NH3 and 0.125 molL−1 in NH4Br. What mass of HCl will this buffer neutralize before the pH falls below 9.00? I GOT THE ANSWER TO BE 0.11g Part 2 IS MY PROBLEMIf the same volume of the buffer were 0.270 molL−1 in NH3 and 0.395 molL−1 in NH4Br, what mass of HCl could be handled before the pH fell below 9.00?

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Homework Answers

Answer #1

No of mol of NH3 in buffer = 0.13*0.105 = 0.01365 mol

No of mol of NH4Br = 0.13*0.125 = 0.01625 mol

pOH = PKb + log(salt+HCl)/(base-HCl)

14-9 = 4.75+log((0.0165+x)/(0.01365-x))

x = No of mol of HCl = 0.0028 mol

mass of HCl = 0.0028*36.5 = 0.1022 grams

similarly

No of mol of NH3 in buffer = 0.13*0.27 = 0.0351 mol

No of mol of NH4Br = 0.13*0.395 = 0.05135 mol

pOH = PKb + log(salt+HCl)/(base-HCl)

14-9 = 4.75+log((0.05135+x)/(0.0351-x))

x = No of mol of HCl = 0.00398 mol

mass of HCl = 0.00398*36.5 = 0.145 grams

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