1) A 0.100 mole quantity of a monoprotic acid HA is added to1.00 L of pure water
ID: 691113 • Letter: 1
Question
1) A 0.100 mole quantity of a monoprotic acid HA is added to1.00 L of pure water. When equilibrium is reached, the pH ofthe solution is 3.75. What is the value of Ka for theacid HA? 2) A 0.250 g quantity of an acid is titrated with 0.108 MNaOH. It takes 27.22 mL of NaOH to reach the end point. What is theequivalent weight of the acid? 1) A 0.100 mole quantity of a monoprotic acid HA is added to1.00 L of pure water. When equilibrium is reached, the pH ofthe solution is 3.75. What is the value of Ka for theacid HA? 2) A 0.250 g quantity of an acid is titrated with 0.108 MNaOH. It takes 27.22 mL of NaOH to reach the end point. What is theequivalent weight of the acid?Explanation / Answer
The given Acid is HA<------> H+ + A- Initial(M) 0.100M 0.00 0.00 At Equilibrium(M) 0.100-10-3.75 10-3.75 10-3.75 Ka=[ H+ ][A-] / [HA] = ( 10-3.75 )2/ 0.100-10-3.75 =(1.778*10-4)2/ 0.0998 = 3.2*10-8 /0.0998 =3.168*10-7 At Equilibrium THe total number of moles of bothAcidnad Base is equal So The mass of Acid m=0.250 g concentrationC=0.108 M Volume V=27.22 mL Concentration C=[moles/V(ml)]*1000(mL) =[(mass/molar mass)/V(mL)]*1000(mL) Molarmass= (0.250*1000)/ (0.108*27.22) = 85.04 g/mol (Rb molarmass=85.4678 g/mol) = 3.2*10-8 /0.0998 =3.168*10-7 At Equilibrium THe total number of moles of bothAcidnad Base is equal So The mass of Acid m=0.250 g concentrationC=0.108 M Volume V=27.22 mL Concentration C=[moles/V(ml)]*1000(mL) =[(mass/molar mass)/V(mL)]*1000(mL) Molarmass= (0.250*1000)/ (0.108*27.22) = 85.04 g/mol (Rb molarmass=85.4678 g/mol)Related Questions
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