1 for 2700 ml of pure water calculate the initial ph and the...

1. For 270.0 mL of pure water, calculate the initial pH and the final pH after adding 0.032 mol of NaOH. 2. For 270.0 mL of a buffer solution that is 0.235 M in HCHO2 and 0.280 M in KCHO2, calculate the initial pH and the final pH after adding 0.032 mol of NaOH. (Ka(HCHO2) = 1.8×10^−4.) 3. For 270.0 mL of a buffer solution that is 0.305 M in CH3CH2NH2 and 0.245 M in CH3CH2NH3Cl, calculate the initial pH and the final pH after adding 0.032 mol of NaOH. (Kb(CH3CH2NH2) = 5.6×10^−4.)

Transcript

0:00 Hello.

00:01 So in the first question we have given that is the moles of n -a -o -h it is given here of n -a -o -h that is equal to 0 .032.

00:24 And the volume it is given that is the 270 ml.

00:31 So this can be converted into 270.

00:36 By 10 x to the power minus 3 liter.

00:39 So now and we need to find out here the initial and the final ph value.

00:50 So the initial ph, this initial ph, as we know that the pure water, it is neutral for the water.

01:05 Pure water it is there so therefore the ph value it is 0 because this is neutral in nature.

01:12 Now the concentration of nahoh.

01:16 We know that, that is the concentration, it is equal to moles divided by volume.

01:25 Hence, 0 .032 divided by 270 multiplied by 10 raised to the power minus 3.

01:32 And this will become 0 .1185 mole per liter, inverse per liter.

01:41 So now we can find out the value of the ph because any osh concentration we have and the concentration of naoh and osh negative will be the same.

01:50 So this poh will be is equal to minus log osh negative concentration and hence minus log oh it is 0 .1185.

02:03 This is equal to 0 .9263.

02:07 And therefore the ph value will be here 14 minus poh that is equal to 14 minus 0 .9263 and this is equal to 13 .0734.

02:26 So the initial ph is equal to 7 and final ph value it is 13 .07334.

02:34 Now for the next we have some given values and again we need to find out.

02:39 The initial and the final ph value.

02:42 So now as we can see that is the p or k a value it is given here as 1 .8 multiply by 10 raised to the power minus 4.

02:59 Now taking the log here that is minus log k a is equal to minus log 1 .8 multiply by 10 raised to the power minus 4 and this minus log k -a, it is nothing but the p -k -a.

03:16 And this will be the value as 3 .7447 it is.

03:22 So now for the initial ph, we have this ph value, let us say initial i here.

03:34 So this p -k -a plus log salt, salt, divided by acid it is there.

03:44 And we have this value 3 .7447 plus log salt it is given that is kh c o h that is 0 .280 divided by acid that is hc oohoh it is there so this is 0 .235 and hence the ph initially this we will have ph or initial p h it is equal to 3 .8208 so now after adding 0 .03 to mole n a .oh...

Question

Please give Ace some feedback