Question

Calculate the ph of the resulting solution if 29.0 ml of 0.290 m hcl(aq) is added to

Answer 1

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

Answer 2

We have the given reaction as;

$HCl_{(aq)} + NaOH_{(aq)}$ ----> $NaCl_{(aq)} + H_{2}O_{(l)}$

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 $10^{-3}$ moles 

and for NaOH we have, n(NaOH) = (0.0390 L)(0.290 mol/L) = 1.13X $10^{-2}$ moles 

Now, it seems NaOH is in excess, so amount remaining will be; 

1.13 X $10^{-2}$ - 8.41 X $10^{-3}$ = 2.89 X $10^{-3}$ moles 

Now, the total volume will become as  = 0.0390 + 0.0290 = 0.068 L 

So, the concentration of [$OH^{-}$] = 2.89×10ˉ³ mol / 0.068 L = 4.25 X $10^{-2}$ M 

pOH = - log [$OH^{-}$] = -log (4.25×$10^{-2}$) = 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 $10^{-3}$ mol 

n(NaOH) = (0.0190 L)(0.390 mol/L) = 7.41 X  $10^{-3}$ mol 

here we can see, HCl is in excess amount so the remaining will be;  

8.41X $10^{-3}$ - 7.41 X $10^{-3}$ = 1.0 X $10^{-3}$ mol 

Here, the total volume will be = 0.0290 + 0.0190 = 0.0480 L 

So the concentration of [HCl] = 1.0 X $10^{-3}$ mol / 0.0480 L = 2.08 X $10^{-2}$ M

 

Which is = [H⁺] 

So, the pH = - log [$H^{+}$] = -log(2.08X $10^{-2}$) = 1.68

 

Hence, the pH will be 1.63 which is more acidic in nature.