A weak base BOH of concentration 0.02 mol litre−1 has a pH of 10.45. If 100mL of thsi base are mixed with 10mL of 0.1MHCl, what will be the pH (write only integer part) of mixture?
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Solution
The number of millimoles of HCl is 10mL×0.1M=1millimole The number of millimoles of base are 0.02M×100M=2millimole Thus out of 2 millimoles of base, 1 will be neutralized by HCl to form 1 millimole of salt. Thus 1 millimole of base will remain. Hence, logsaltbase=log11=0. Hence, the pOH of the buffer solution will be equal to pKb of base.
Initial concentration (M)
Equilibrium concentration (M)
BOH
0.02
0.02-x
B^+
0
x
OH^-
0
x
pH=10.45,pOH=14−10.45=3.55 Hence, x=[OH−]=10−pOH=10−3.55=2.818×10−4 The equilibrium constant expression is K=[B+][OH−][BOH] Substitute values in the above expression. K=(2.818×10−4)×(2.818×10−4)0.02−(2.818×10−4)=3.97×10−6 Hence, pKb=−logK=−log3.97×10−6=5.40 pOH=pKb=5.40 pH=14−pKb=14−5.40=8.598