A calculate the ph in the titration of 1000ml of 10m hcl...

a. Calculate the pH in the titration of 10.00mL of 1.0M HCl with 1.0M NaOH after the addition of 2.0mL NaOH? b. Calculate the pH in the titration of 10.00mL of 1.0M HCl with 1.0M NaOH after the addition of 10mL NaOH? c. Calculate the pH in the titration of 10.00mL of 1.0M HCl with 1.0M NaOH after the addition of 11mL NaOH? d. A student wants to prepare 250 mL of a standard solution of sodium carbonate to titrate an HCl solution that is thought to be approximately 0.5 M. In the titration she will be using a 25 mL pipette for the Na2CO3 and a 50 mL burette for the HCl. What mass of sodium carbonate should she weigh out? include any assumptions you make in doing this calculation. e. Ammonium ion in lawn fertiliser is analysed in the following way. 1.00 g of the fertiliser is dissolved in water. 30.00 mL of 0.7500 M NaOH are added to the solution which is then boiled to remove all NH3. 12.60 mL of 0.8555 M HCl are needed to titrate the excess NaOH. What is the % of NH4+ in the fertiliser sample? f. An impure sample of Al(OH)3 weighing 1.265g is dissolved in acid and the Al3+ is eventually precipitated as pure Al2O3. The dry precipitate is found to weigh 0.4278 g. What is the % of Al in the sample?

Transcript

00:01 So let's first write down the chemical reaction.

00:03 So we have hydrochloric acid plus sodium hydroxide.

00:09 And this will make, of course, sodium chloride, the salt, plus water.

00:17 This is already balanced.

00:18 And so we can just go right ahead and say that first we have 10 milliliters of hydrochloric acid at 1 .0 molar concentration.

00:31 So the number of moles here would be equaling to 0 .010 liters.

00:37 This would be multiplied by 1 .0 molars, and we are left with 0 .010 moles of hydrochloric acid.

00:50 Now, for part a, we are adding sodium hydroxide.

00:55 We're adding sodium hydroxide for part a, b, and c.

00:59 So for part a, we have two milliliters at one molar concentration.

01:12 And so we can actually multiply these two, and we get 0 .00 -20 moles of sodium hydroxide.

01:24 And the mole ratio is one to one for these.

01:29 So we can say that then the concentration of hydrogen ions, or in this case, protons, would be equaling to 0 .010 moles of hydrochloric acid minus 0 .020 moles of sodium hydroxide.

01:52 And this will all be divided by 10 milliliters plus 2 milliliters or 0 .012.

02:01 To liters.

02:04 And so this gives us a concentration of 0 .667 molar.

02:10 And we can find the ph.

02:13 So the ph would be equaling to the negative log base 10 of the hydrogen ion concentration.

02:24 And this is equaling that 0 .176.

02:27 So very, very acidic still.

02:32 For part b, we have sodium hydroxide.

02:37 And again, hydrochloric acid.

02:41 But here, they're both 10 milliliters at 1 .0 molar concentration.

02:51 Again, they have a 1 -to -1 mole ratio as deduced by the chemical reaction.

02:58 And so we can say that this would be perfectly balanced, or we have a ph equaling to 7.

03:06 So this would truly just result in salt water.

03:11 So would it be basic? would it be acidic? and then for part c, we actually are adding more sodium hydroxide than necessary.

03:23 So it's actually better to figure out the concentration of hydroxide ions.

03:31 So this would be equal to 0 .01 liters or 11 mil liters of sodium hydroxide at 1 .0 molar.

03:45 This would be then minus the 10 mil liters, 0 .010 liters multiplied by 1 .0 molar.

03:55 And then this would be divided by 0 .01 liter plus 0 .01 .1.

04:03 0 .0 .0 .21 liters.

04:09 Giving us 0 .021 liters, giving us a concentration of 0 .0476 molar.

04:18 Now, from this, we can say that then the p .o .h would be equaling to the negative logarithm, base 10, again, of the hydroxide ion concentration.

04:30 This is giving us then 1 .3 approximately.

04:36 And so we know that ph plus p .o .h would be equaling to 14.

04:43 And so we can say the ph would be equalling to approximately 12 .7, or 14 minus 1 .3, 12 .7.

04:56 For part d, we have hydrochloric acid again.

05:02 However, now we have 50 milliliters or 0 .050 liters at 0 .5 molar concentration.

05:16 This is giving us our 0 .025 molar.

05:21 And so here we are reacting hydrochloric acid with sodium carbonate...