Question

How many mls of 0.125 M HCI must be added to 65.5 mls of 0.234 M NH3 to completely neutralize the NH3? 65.5 369 34.9 246 O 123

Answer 1

The answer is

Nho reacts with Hel to give Nitycl NH3 + HCl - > HyCl -> volume of NH 65.5ml - 65:5X10-3L (:1m)=1013L) = 0.0655 L. conc. of NH = 0.23411 we know that, Molarity & No. of moles solut vor of sonin .NO.04 moies 04 NH, Molanity x Vol 04 ccone) som in = 0.2341V X0-0655 20.015 moles From the balanced chemical equation & it is clear that 1 mole of NH₃ is neutralised by = 1 mole of Hali 0.015 mole of NH3 is neutralised by = 0.015 mole of He : No. of moles of Her needed to neutralise 65.5 m/ 0.234 M NN3 -0.015 more

conc. of Hel solution available = 0.125M we know, Molarity = no. of moles of Hal volume of soin in L 2. volume of = no. of moles of Soin in L HCl Molasity 0.015324 moles 0.125M - 0.123 L - 0.123 x 103 m C:.1 L=103 = 123 m) o)) : 123 ml of 0.125 M Helmust be added to 65.5 ml of 0.234 M) NH₃ to completely neutralise the NHA