A solution is prepared by adding 50 mL of 0.03mol/L HCl to?

Question

40 mL 0.04mol/L NaOH. Calculate the pH of the solution.

Show your work.

Answer 1

HCl + NaOH === NaCl + H2O
0.0015 , 0.0016 , 0 , 0 at start of reaction (moles)
0, 0.0001 , 0.0015 , 0.0015 at end of reaction(moles)

every mole of water contributes about 10^-7(ten to the power -7)
moles of OH- and H+ . thus their contribution to the pH is neglected when there is a considerable amount of acid or base present i.e more than 10^-5 moles/litre (not moles but CONCENTRATION)
Thus concentration of NaOH = 0.0001/90*1000 moles/litre{total volume = 50+40 ml} = 0.001 .
as NaOH is a strong electrolyte like any other sodium compound we can assume complete ionisation . thus[OH-]=[NaOH] = 0.001. now pH = -log[H+]=14-(-log[OH-])= 14+log[OH-] =14+log[0.001]=11
thus the pH=11

Answer 2

First put everything into moles (0.03 mol/L HCL * .050 L) & (0.04 mol/L NaOH * .040 L). Now divide your moles by total volume (0.0015 mol HCL/ 0.090 L) & (0.0016 mol NaOH/ 0.090 L). Now, the one with the lesser amount will be the one completely used (16.67 x 10^ -3 M HCL < 17.78 x 10^ -3 M NaOH), so you go by the reactant that will have "leftovers", in this case, NaOH.

Now you set up your table
Base + H20 <==> OH + HB
17.78*10^ -3 -- 0 0
-x -- +x +x
(17.78*10^ -3)-x -- x x

Kb = (x*x)/ (17.78*10^ -3) = (value of Kb should be in your book or previously given to you).

Now, you just have to solve for "x". ( square root of Kb/ [17.78*10^ -3]). x = [OH-].

Now, -log [OH-] = pOH. Then 14- pOH = pH.

Good luck!!

Answer 3

HCl: No. of moles = 0.03 * (50/1000) = 0.0015 moles
NaOH: No. of moles = 0.04 * (40/1000) = 0.0016 moles

HCl + NaOH --> NaCl + H2O
Since the mole ratio is 1:1, based on the above calculations, there would be an excess of NaOH.

Concentration of NaOH
= moles / total volume
= (0.0016-0.0015) / ((50+40)/1000)
= 0.001111 mol/L

pH = 14 - (-log[OH-]) = 14-(-log(0.001111)) = 11.0 (to 3 sig fig)

Answer 4

HCl + NaOH ----> NaCl + H2O 1 M of HCL will react with 1 M of NaOH to form 1 M for NaCl. since we have only 1.6 M of HCl it would form 1.6 M NaCl. (althhough 1.8 M of NaOH is present... it does not have suffficient HCL to react so 0.2 M of NaOH would still remain as it is)