<p>Hey, the answer's online but I can't decipher the symbols so could someone explain the answer to this question in words? Thanks~</p>
<p>6) A 0.500-gram sample of a weak, nonvolatile acid, HA, was dissolved in sufficient water to make 50.0 milliliters of solution. The solution was then titrated with a standard NaOH solution. Predict how the calculated molar mass of HA would be affected (too high, too low, or not affected) by the following laboratory procedures. Explain each of your answers.</p>
<p>(a) After rinsing the buret with distilled water, the buret is filled with the standards NaOH solution; the weak acid HA is titrated to its equivalence point.</p>
<p>(b) An indicator that changes color at pH 5 is used to signal the equivalence point.</p>
<p>(c) An air bubble passes unnoticed through the tip of the buret during the titration.</p>
<p>a) you should always add a bit of whatever you're titrating with (in this case, NaOH) into the buret, rinse the buret with the NaOH, then dump it out and put in the measured NaOH. If you don't, there's gonna be water in the buret as you titrate, which means that when you finish the titration with, say 50mL of base, you'll actually have maybe 49mL. Thus, you use more base to titrate = more moles OH- than you really need to equalize solution = more moles H+ in HA = higher MM (since is moles/L). Thus, MM is too high.</p>
<p>b) Because this is a weak acid, it will have a conjugate base that is a stronger base than water, and at the equivalence point there will be mainly conjugate base ions, thus EQ will be slightly basic. Thus, a indicator that changes color in the acid range will not work, because it'll change colors too quickly, meaning you think equilibrium is reached before it is actually reached = less mL base used than really neeed = fewer moles H+ in HA = lower MM. Thus, MM is too low.</p>
<p>c. Same explanation as part a. Since air has no H+ ions, you'll get more volume base used than actually necessary = more moles OH- calculuated = more moles H+ calculated = too high of a MM.</p>
<p>thanks, except that the answers are just the opposite of the above, according to the AP exam answer guide.... (a) is supposed to be too low, (b) too high, and (c) too low... which is why i was slightly confused</p>
<p>1996 AP Chem Free Response... i think i get b) though... since NaOH is a strong base, the pH should be greater than 7 after titration, so having a pH = 5 means not enough base was added, meaning the volume is too low. Calculated moles for NaOH will then be too low, and since NaOH mol = HA mol, then calculating molar mass of HA means the 0.500 grams will be divided by a smaller number, so MM too high.</p>
<p>but that doesn't matter, because at the EQ, there is no OH- from NaOH. It's only the conjugate base, which is basic. And before, pH rises pretty slowly, and is still acidic because the OH- gets paired up with H+. Urg! I've totally sucked at chem today!</p>
<p>actually, ebony tear, what you said makes me get it now, haha, weird as that sounds... </p>
<p>in (a), since the water dilutes the NaOH and the OH ions, then you add more NaOH than needed, meaning more volume. Since you presumably know the molarity of NaOH, then mol/L (a larger "L" value) means mol of NaOH is greater than expected, and mol NaOH = mol HA, so MM of HA = g/mol and 0.5 / (a larger mol value) = too low of a MM! </p>