1) The flask shown here contains 10.0 mL of HCl and a few drops of phenolphthale
ID: 474451 • Letter: 1
Question
1) The flask shown here contains 10.0 mL of HCl and a few drops of phenolphthalein indicator. The buret contains 0.280 M NaOH.
2) A patient is suspected of having low stomach acid, a condition known as hypochloridia. To determine whether the patient has this condition, her doctors take a 17.00 mL sample of her gastric juices and titrate the sample with 0.000332 M KOH. The gastric juice sample required 8.50 mL of the KOH titrant to neutralize it. Calculate the pH of the gastric juice sample. Assume the sample contained no ingested food or drink which might otherwise interfere with the titration.
3) A monoprotic acid, HA, is dissolved in water:
4) During exercise when the body lacks an adequate supply of oxygen to support energy production, the pyruvate that is produced from the breakdown of glucose is converted into lactate. High lactate levels can lead to acidity in the muscle cells as some of the lactate hydrolyzes to lactic acid. The dissociation of lactic acid to lactate is shown in the reaction below. Lactic acid has a pKa of 3.86.
The flask shown here contains 10.0 mL of HCl and a few drops of phenolphthalein indicator. The buret contains 0.280 M NaOH What volume of NaOH is needed to reach the end point of the titration? Number 16 mL NaOH What was the initial concentration of HCI? Number 1.4 M HCl 10 30 35 45 Reset 10 15 20Explanation / Answer
1) NaOH +HCl --------------> H2O + NaCl (1:1 ratio )
The solution turned pink when 16 mL of NaOH was added to HCl
16 mL * 0.280 M = 4.48 mmoles of NaOH were added
at equivalence point , mmoles of HCl = mmoles of NaOH
4.48 mmoles of HCl = 10mL * X M of HCl
let "X" be the molarity of HCl
X = 0.448 M = intial HCl concentration
2) KOH +HCl ------------> KOH + H2O (1:1 ratio)
mmoles of KOH used = 0.000332 M * 8.50mL = 2.822*10-3 mmoles
mmoles of HCl used = 2.822*10-3
HCl has 2.822*10-3 mmoles of H+ ions
[H+] = ( 2.822*10-3 mmoles) / 17 mL = 1.66*10-4 M
PH = - log10 [ H+] = 3.78
Since the PH of his gastric juices is less than 4 we can say that patient is not suffering from hypochloridia.
3) Use Henderson-Hasselbalch equation,
PH = PKa + log ([A-] / [HA] )
PH = -log10[H+] = - log (3*10-4) = 3.52
3.52 = PKa + log (3*10-4 / 0.180)
PKa = 6.298 = 6.3 approx
4) PH = PKa + log ([lactate] / [lactic acid])
3.01 = 3.86 + log ([lactate] / [lactic acid])
-0.85 = log ([lactate] / [lactic acid])
[lactate] / [lactic acid] = 10-0.85 = 0.1412
fraction lactic acid is same as above
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