Question

What would the pH of a solution that contains 25.1 mL of 0.50 M ammonia? What would the pH be if 2.78 mL of 0.0471 M HCl was added to the ammonia solution? Please show all work!!!

Answer 1

0.025 L Given Molarity of NH3 = 0.500 M Volume of NH3 = 25.10 ml = We know, Moles = Molarity (M) x Volume (V in L) Moles of NH3 = 0.500 MX Molarity of HC1 = 0.047 M 0.0251 L = 0.01255 mol 1.Initial pH of solution or at 0.0 ml Base addition Acetic acid is a weak acid, it dissociate partially into acetate and OH- in aqueous media as shown below NH3(aq) + H2O (1) <-----> NH4+ (aq) + OH- (aq) Initial 0.5000 0 Change Equilibrium 0.500 - X + X Assume x is small then 0.500 - x will be = 0.500 Given Kb = 1.80E-05 Weak acid dissociation constant Kb = (NH4+)(OH-)/(NH3) 1.80E-05 = x*x+( 0.500 - x) 1.80E-05 x*x+ 0.5 x^2 1.80E-05 * 0.500 sox ^2 = 9.00E-06 0.00300 but from ICE table x = [OH-]; so [OH-] = 0.0030000 M but pOH = -log [OH-] = -log[ 0.00300000 = 2.52288 we Know PH + pOH = 14 or pH = 14-pOH = 14 - 2.5229 = so pH of solution = 11.5 11.48 2. After addition of 2.8 ml base Given Molarity of NH3 = 0.500 M Volume of NH3 25.10 ml = 0.025 L We know, Moles = Molarity (M) x Volume (V in L) Moles of NH3 = 0.500 Mx 0.0251 L = 0.01255 mol 0.0028 L Volume of HCI= Molarity of HC1= Moles of HCI= 2.8 ml or 0.047 M 0.047 mol L-1* 0.0028 L = 0.00013 mol Kb = 1.80E-05;pKb = log (Kb) = -log (1.80E-05 ] = 4.7447 we Know pKa+pKb = 14 or pKa = 14-pka = 14 - 4.7447 = 9.255 When H+ is added to Weak Base, it reacts with and produce Conjugate acid We know both acetic acid and aceate act as a buffer system. The ICE table is shown below. NH3(aq) + H+ (aq) <-----> NH4+ (aq) Initial 0.01255 0.00013 0 .000 Change -0.00013 -0.00013 +0.00013 Equilibrium 0.01242 0.00013 0 + Henderson equation for buffer solution is pH=pKa + log [Base/Acid] Here Base is NH3 and acid is NH4+ so pH =pKa + log[Base/acid] =pKa + log[NH3 /NH4+] = 9.255 log [ 0.01242 0.000131 = 9.255 log [95.0508 ] 9.255 1.9780 11.23 + +