A 111.0 mL sample of 0.105 M methylamine (CH3NH2;Kb=3.7×10−4) is
titrated with 0.265 M HNO3. Calculate...
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Chemistry
A 111.0 mL sample of 0.105 M methylamine (CH3NH2;Kb=3.7×10−4) istitrated with 0.265 M HNO3. Calculate the pH after the addition ofeach of the following volumes of acid:
22.0 mL
Express the pH to two decimal places.
44.0 mL
Express the pH to two decimal places.
66.0 mL
Express the pH to two decimal places.
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Chemical equation for the acid base reaction is CH3NH2aq HNO3aq CH3NH3aq NO3aq 1 mol 1 mol 1 mol 1 mol moles of CH3NH2 initially taken MxV 0105 molL x 0111 L 001166 mol a When 220 mL 0265 M HNO3 is added Moles of HNO3 added MxV 0265 molL x 0022 L 000583 mol HNO3 CH3NH2aq
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