We have the given reaction as;
[tex]HCl_{(aq)} + NaOH_{(aq)}[/tex] ----> [tex]NaCl_{(aq)} + H_{2}O_{(l)}[/tex]
Answer A) The pH will be 12.36,
We have to convert the concentrations of HCl and NaOH into moles,
So we have, n(HCl) =
(0.0290 L) X (0.290 mol/L) = 8.41X [tex]10^{-3}[/tex] moles
and for NaOH we have, n(NaOH) = (0.0390 L)(0.290 mol/L) =
1.13X [tex]10^{-2}[/tex] moles
Now, it seems NaOH is in excess, so amount remaining will be;
1.13 X [tex]10^{-2}[/tex] - 8.41 X [tex]10^{-3}[/tex] = 2.89
X [tex]10^{-3}[/tex] moles
Now, the total volume will become as = 0.0390 + 0.0290 = 0.068 L
So, the concentration of [[tex]OH^{-}[/tex]] = 2.89×10ˉ³ mol
/ 0.068 L = 4.25 X [tex]10^{-2}[/tex] M
pOH = - log [[tex]OH^{-}[/tex]] = -log (4.25×[tex]10^{-2}[/tex])
= 1.37
Hence, pH = 14 - pOH = 14 - 1.37 = 12.6
So the pH of the
solution will be 12.6 which is basic in nature.
Answer B) The pH will be 1.68
Now, for the given concentration we need to find moles for HCl and NaOH also;
n(HCl) = (0.0290
L)(0.290 mol/L) = 8.41 X [tex]10^{-3}[/tex] mol
n(NaOH) = (0.0190 L)(0.390 mol/L) = 7.41 X [tex]10^{-3}[/tex] mol
here we can see, HCl is in excess amount so the remaining
will be;
8.41X [tex]10^{-3}[/tex] - 7.41 X [tex]10^{-3}[/tex] = 1.0 X
[tex]10^{-3}[/tex] mol
Here, the total volume will be = 0.0290 + 0.0190 = 0.0480 L
So the concentration of [HCl] = 1.0 X [tex]10^{-3}[/tex] mol
/ 0.0480 L = 2.08 X [tex]10^{-2}[/tex] M
Which is = [H⁺]
So, the pH = - log [[tex]H^{+}[/tex]] = -log(2.08X [tex]10^{-2}[/tex])
= 1.68
Hence, the pH will be 1.63 which is more acidic in nature.
a) when 39.0mL of 0.290M NaOH(aq) is added
1- firs we need to get the number of moles (n) for both HCl and NaOH :
when n (HCl)= volume * molarity
= (0.0290 L)(0.290 mol/L)
= 8.41×10ˉ³ mol
and when n (NaOH) = volume * molarity
= (0.0390 L)(0.290 mol/L)
= 1.13×10ˉ² mol
moles of NaOH remaining = 1.13×10ˉ² - 8.41×10ˉ³
= 2.89×10ˉ³ mol
2- then we will get the total volume:
total V = 0.029 L + 0.039L
= 0.068 L
3- we calculate the [OH-] :
[OH-] = moles / volume
= 2.89×10ˉ³ mol / 0.068 L
= 4.25×10ˉ² M
4- calculate POH from the value of [OH-] :
when POH = -㏒[OH-]
= -㏒(4.25 x 10^-2)
= 1.37
5- Calculate the PH from the value of POH:
when PH + POH = 14
∴ PH = 14-1.37 = 12.63
b) when 19.0mL of 0.390M NaOH(aq) is added :
1- firs we need to get the number of moles (n) for both HCl and NaOH :
when n (HCl)= volume * molarity
= (0.0290 L)(0.290 mol/L)
= 8.41×10ˉ³ mol
n(NaOH) = volume * molarity
= (0.0190 L)(0.390 mol/L)
= 7.41x10ˉ³ mol
moles of HCl remaining = 8.41 x 10^-3 - 7.41 x 10^-3
= 1.0×10ˉ³ mol
2- then we will get the total volume:
total V = 0.0290 + 0.0190
= 0.0480 L
3- we calculate the [HCl]=[H+] :
[H+] = moles / volume
= 1.0×10ˉ³ mol / 0.0480 L
= 2.08×10ˉ² M
4- finally we calculate the PH from the [H+] value:
when PH = -㏒[H+]
= -㏒ 2.08 x 10^-2
= 1.7
