Answer to Question #91281 in General Chemistry for Morgan Washington

Question #91281
what volume of 0.125 M HCL is needed to titrate 50.0 ML of 0.0500 M NH3 solution. ML
1
Expert's answer
2019-07-01T03:54:33-0400
"HCl(aq) + NH_3 (aq) = NH_4Cl(aq)"

At the equivalence point, the number of moles of HCl added is equal to the initial number of moles of NH3, because the analyte is completely neutralized.


"n(NH_3)=c\\times V = 0.0500 \\frac{mol}{L} \\times 50.0\\times 10^{-3} L = 0.0025 mol"

"n(NH_3) = n(HCl)"

"n(HCl) = 0.0025 mol"

"c = \\frac{n}{V}"

"\\therefore V = \\frac{n}{c} = \\frac {0.0025 mol}{0.125\\frac{mol}{L}}= 0.02 L = 20.0 mL"



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