Question

21. -/0.1 points 0/4 Submissions Used Calculate the pH of each of the following strong acid solutions. (a) 0.00476 M HCI pH = (b) 0.850 g of HI in 41.0 L of solution pH = (c) 14.0 mL of 2.10 M HCl diluted to 1.50 L PH = (d) a mixture formed by adding 60.0 mL of 0.00416 M HCl to 57.0 mL of 0.00234 M HI pH =

Answer 1

Ang @ 2 PH= - 10910 [H"]. 0.00476 M HCl Pha -logio (0.00476) = 12.32 Ang 0.85 g HI in uil Solution mole of HI: 0.85 - 6 128 [HI] = 0.85 128 x 41 i = - 0.00016) M 0 PH= -logo (0.000161) - 3.79] Any 14 mi of 2.1 M HCl diluted to 1.5L mole of Hc Molarity x Volume = 9.1 x 0.014 = 0.0294 moi [MCI ] = 6:0284 = 0.0196 M PHE -10910 (0.0196) = 1.70 By Gomi of o.o0u16 M HCl +57m1 of 0.00234 M HI M, V + M2V2 60X0.00u16 + 57X0:00 234 Go +57 [ht Iner = Vit V2 [Hit het = 0.00327 M Plt = -109,0 (0.00327) = 2.48 Ang s ed with