Question

Consider the reaction NH3 +HCL=NH4Cl, where 100 ml of 5M NH3 and 100ml of 5M of HCL are combined in the same beaker. Calculate the pH at equilibrium.

Answer 1

No of moles of NH2 -100 mLx x Noof moles of HCI added =100 mL ×--× ICE table 1000 mL L -IL -X 5 nol-0.500 mol NH, + H+-> NH; I(mol)000 0.500 C(mol) 0500 -0.500 0500 E (mol) 00 At equival ence point, pH of the weak base Vs strong acid can be calculated by the hydrolysis of conjugate acid of the weak base Total volume = (100 +100) mL = 200 mL = 0.200 L Concentration of NH4+- 0.500 0.500 mol 0.200 L 2.50 MM Equation representing the ionization of NH, is as follows Baseionizati on constantK, of NH3-1.8x10 Kw ,-- 1.0x10- 18×10 Acidi onization constant Ka of NH 5.6×10-10 = ) 2.50 Initi al concentrati on M Change (M) Equilibrium concentration (M) : 2.50-х Aadi oni zation constantX,ÎH31H3의 NH, х 2.50-х 5.6x10-10- From the equilibrium table, xH3.74x105 M pH--log[H,o]log (3.74x10-3) 443