▼ Part C Calculate the equilibrium concentrations of Na+, Cl ,H+, CHOnd HCHO2 when 50.0 mL of 0.2...
▼ Part C Calculate the equilibrium concentrations of Na+, Cl ,H+, CHOnd HCHO2 when 50.0 mL of 0.20 M HCl is mixed with 50.0 mL of 0.20 M NaCHO Express your answers using two significant figures. Enter your answers separated by commas. Request Answer
▼ Part C Calculate the equilibrium concentrations of Na+, Cl ,H+, CHOnd HCHO2 when 50.0 mL of 0.20 M HCl is mixed with 50.0 mL of 0.20 M NaCHO Express your answers using two significant figures. Enter your answers separated by commas. Request Answer
50.0 mL ×0.2011 Concentration of CHO, CHOİ |-Na CE102] = = 0.10M (50.0+50.0) mL Concentration of Na. [Na·J-[NaCH02-500 mL 0 20M (50.0+50.0) mL = 0. 1014 」 50.0 mL ×0.2011 (50.0+500) mL 50.0 mL (50.0+50.0) mL -10 Concentrati on of H H HC1- = 0. 1014 0.20 M Concentrati on of CIClHCI- Concentration ofar, [a-J-[Ha]= = 0 1014 ICE table I(M) 0.10 C(M): -0.10-0.10 E(M): 0 0.10 +0.10 0.10 ICE table I(M)0.10 C(M) E(M): 0.10-x Acidionization constant, K. - HCHO 1.8×10-4 0.10-х Since HCHO is a weak acid we can assume that (0.10-x)M = 0.10 M 1.8x10-4- 0.10 x-J(1.8×10-4 )×(0.10) x = 4.2×10-3 M According to the equilibrium table H+ |=(x)M= 4.2×10-3 M CHOg (x)M= 4.2×10-3 M [HCHO:Jr( 0. 10-z)11-(0.10-4.2x10-3)M = 0.10 M