Answer to Question #179039 in General Chemistry for Sum

Question #179039

Consider the titration of 40.0 mL of 0.0600 M (CH3)2NH (a weak base; Kb = 0.000540) with 0.100 M HCI. Calculate the pH after the following volumes of titrant have been added: (a) 0.0 mL (b) 6.0 mL (c) 12.0 mL (d) 18.0 mL


1
Expert's answer
2021-04-07T11:37:46-0400

B) after 6.0 ml

(Base = Salt ) which show below


"\\frac{M_1\u00d7V_1}{V_1+V_2}= \\frac{M_2\u00d7V_2}{V_1+V_2}"


"(Base)=\\frac{0.06\u00d740}{40+6}=0.0522M"


"(Acid)=\\frac{0.1\u00d76}{40+6}=0.013M"


Base left = "0.0522-0.013=0.0392 M"


"POH=Pkb+ log(\\frac{salt}{base})"


"POH=3.36+ log(\\frac{0.013}{0.0392})"


"PH=14-POH=14-2.88=11.12"


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