A buffer solution of baking soda and washing soda was added to water and diluted. The pH was tested at approx. 10 on universal pH paper. Household ammonia and vinegar were added in separate buffer solutions and no pH changes were noted.
What ions were present in the buffer solution? Identify the weak acid and its conjugate base.
Thanks for your help.
2006-12-22 08:57:32 · 5 answers · asked by luc_allmon 2 in Science & Mathematics ➔ Chemistry
Write the equilibrium equation (expression) for your buffer.
2006-12-22 08:58:30 · update #1
Baking soda is NaHCO3. Washing soda is Na2CO3. Ions in the buffer solution were HCO3-, CO3=, and OH-. The weak acid is HCO3-; the conjugate base is CO3=. Because a weak acid and a weak base were added to this strongly alkaline buffered system, no pH changes would be expected.
2006-12-22 09:36:33 · answer #1 · answered by steve_geo1 7 · 0⤊ 0⤋
Formula for baking soda (sodium bicarbonate): NaH(CO3) (weak acid)
Formula for washing soda (sodium carbonate): Na2(CO3) (weak base)
Formula for ammonia: NH3 (a weak base)
Formula for vinegar (acetic acid): HC2H3O2 (a weak acid)
Background knowledge: I already labeled the acids and bases. Acids donate H+, while bases accept H+. Salts (ionic compounds like vinegar and the two carbonates) dissociate (ionize) in water, therefore the sodiums of the carbonates and some of the H+ from the vinegar are ions dissolved in the water.
Now put the weak acid of the buffer together with the weak base (NH3) in one equation, and then write another equation with the weak base of the buffer with the weak acid (acetic acid). Write out all the ions that are formed for the reactants in water, and then put the H+ where there is a negative charge on the product side.
Can you take it from here?
2006-12-22 09:30:52 · answer #2 · answered by teachbio 5 · 0⤊ 0⤋
___HNO2 <-- --> H+ + NO2- 2.fifty 9*10^-2_____0____0__ 2.fifty 9*10^-2 -X___X____X__ Ka = 4.5*10^-4 = X^2 / (2.fifty 9*10^-2 - X) ... X = ... = [H+] *** buffer: [H+] = Ka*[HNO2] / [NaNO2] = = 4.5*10^-4 * 2.33*10^-2 / 2.fifty 9*10^-2 = ... *** hydrolysis: NaNO2 -----> Na+ + NO2- 2.fifty 9*10^-2 __/_____0__ _0_________/___2.fifty 9*10^-2 NO2- + H2O <-- --> HNO2 + OH- 2.fifty 9*10^-2 ________0_____0_ 2.fifty 9*10^-2 -X______X_____X_ Kb = Kw/Ka = 10^-14 / (4.5*10^-4) = X^2 / (2.fifty 9*10^-2 -X) .. X = ... = [OH-] --> pOH ---> pH = 14 - pOH
2016-12-15 06:23:36 · answer #3 · answered by ? 4 · 0⤊ 0⤋
I think that's what they use to make Jim Beam.
2006-12-22 09:15:44 · answer #4 · answered by Dumb Dave 4 · 0⤊ 0⤋
eachonet or whatever her/his name is practically spelled it out for you. I think that answer is the best.
2006-12-22 10:06:21 · answer #5 · answered by I <3 Animals 5 · 0⤊ 0⤋