Question

13. Calculate the pH when 35mL of 0.5M ammoni 0.2M hydrochloric acid. HICI= 208,2 nmole NH32 35X0.5= 17.5 moles pOH = pk bt log bolt=4,76+ log bolt = 4.23 salt base in 13 ,5 NH3t HCl - NHACI Initial 17.5 u final 13,5 - wean bast salt Basic Buffer pH = 14-4.23 = 9.76 14. Sketch the titration curve when 35mL of 0.5M ammonia (K5=1.8x10^-5) is titrated with 0.2M hydrochloric acid.

Answer 1

13 answer is

Given base NH Acid HCl Houmber of moles = Molarity X Voluwe Cin L) : Ny moles V = 35 ml = 35 X 10 L M = 0.5 M no (95x16" ) (0.5 mein) = 11.5 xvő?mol Hcl moles U= gome = qox10 L, M = o.am ne (20x152) Cord molai) = 4 x 10? mol Nth + HCl - Initial: 17.5x10 4x16? Change: - 4x103 -4X15" NHACI : 0 +4x10-3 Finally: 13.5X163 0 4x10? A NH₃ and NHACI is present in that solution. ire. Basic buffer is formed .

Basic buffer = pas + log ? pOH = pk + lon number of moles of salt huwbex of moles of Base s-log Kyo -109 i 118x16-5) = -10g (1-3) -10g (135) =-0.255 - (-5) log 40 5-0.255 pky = 4.th pe pol - = 44.71 + lo 4X103 [ 13.5 X163 = 4.74 + log co.2963) 54.74 - 0.528 = 4.21 - PH + por = 14 pH = 14 - por pH = 14-4:21 =9.29 pH = 9.8