Calculating the Resulting pH

Calculating the pH of the resulting solution when Hydrochloric Acid is mixed with Sodium Hydroxide. First step is to calculate which solute is in excess and how much of the excess reagent is remaining after neutralisation has occurred.

From this we then were able to use the equilibrium constant for water to determine the concentration of hydrogen ions were remaining in solution and then from that calculate the resultant pH of the solution.

I hope the example helps :)

[himeshsingh]

I can surely say that the unrest caused due to other confusing videos for the topic was very well helped by the tutorial. Thanks a lot sir, this is possibly the best one for this topic. (I just hope I remember all of this during the online exam)

[DisneyPlayhousetv]

One day you will tell me who you are so I can leave everything to you in my will for beautifully explaining every single thing on this entire planet.

[Jison_Golosinda]

I been struggling on understanding analytical chemistry but when i see this one, it helps me a lot.

[vetyancie]

thank you so much for the explanation. I was confused with the formulas and the order to use them. thank you so much!!! great work

[shimshonboreri]

super helpful! I cannot thank you enough!

[gegegege]

you are a literal life saver i love you

[PeterPan-xkes]

Or after getting the pOH of (OH), you can also use the formula : 14= pOH +pH. same thing what you did in the video. Thanks for sharing

[pavanvasista]

dude u are so best, explains easily .... u will find the sucess fast

[rich]

my savior. my chem teacher lowkey does not teach

[faithwangui]

Lovely and neat explainaton. Thank you

[mannyalexsalazar]

How would the calculations differ if the ratios in the reactant part of the equation were different? Say if it were a 1:2 ratio or something like that?

[Max-hedq]

Great but this is assuming that the reaction is straight forward, what if we had an equilibrium situation and thus you would have some hcl left and some naoh left.

[buchanan]

Just have one problem 10^-14/6.97*10^-3 =1.43*10^-18 then pH should be 17.8 and I know it's over 14 but that should be the answer...im not sure how you got 1.435*10^-12

[lilialazar]

wouldn't it be faster if you did -log(6.97x10^-3) and then subtract that from 14
-log(6.97x10^-3) = 2.156786104
pH= 14 -2.156786104
= 11.8432139

[scar-let]

Thank u so much😍
It was quite helpful

[trevorh]

Thank you so much! This was really helpful

[deepeshpatil]

According to me
NaOH should be the limiting reagent.

[marcowong]

I got a question. I am planning to prepare a 1-litre solution with a final pH of 11. 0.8 grams of calcium hydroxide (Ca(OH)2) must be included in the solution. I want to know how much pH 7 water and how much concentrated 65% nitric acid I should use to achieve the desired pH of 11 in the final 1-liter solution.

[sonal]

superb video bro....this video help me a lot... thnx

[cheesywiz]

Thank you so much!!!! This was really helpful
But I was wondering ....
When do substitution reactions actually occur? like I think of HCl as being a 'happy couple' with an H+ and a Cl- ( a plus and a minus )
Now why would Cl- for example ... cheat on H+ and go for Na+ instead? Is there a general procedure to figure out when and how a substitution reaction will take place or do I have to learn them individually ...
anyways thanks again for the video