Question

A 30.0 ml. stock solution of 0.447 M ammonium nitrate, NH.NO3, is mixed into a beaker containing 220.0 mL of water. What is the nitrate ion concentration in the resulting solution? 0708 00610M DO

Answer 1

300ml and 0.497m of NHINO, Pain mixed with zzomi of water M, y = M2V V = 30 ml V2= 220 +30 = 250m) M, = 0.447 M NH No, molarity = Normality 10.07MX30ml = M2X256 m) Final solution molanty (M2)=0.0536 M The resulting solution witrate pon concentration in 0.0536m 1 here concentration of whinag = Concentration Nos same. Cor) . The number of moles of whanoz stocle solution is - MXV ne Bolxo-u47 mol/ looo TOGO - = 0.0134 mol There 0.0134 mol of NH qnoz in 220ml of water so The final concentration molarity ) of whywas in = 0.0134 mol y looo here Volume = 30ml + 2zomi (total) = 0.0536 M " - 2somi Concentration of why nos = concentration of Noz 250 L the nitrate ion concentration = 0.0536 mg