Question

Calculate the pH of a solution made by mixing 14.5mL of 0.15M HCl with 5.3mL of 0.10M NaOH. Enter your answer to two places after the decimal.

Answer 1

Answer: -  
pH = 1.08

----------------------------------------------------

Given: -

Volume of HCl = 14.5 ml = 0.0145 L

Molarity of HCl = 0.15 M = 0.15 mol/L

Volume of NaOH = 5.3 ml = 0.0053 L

Molarity of NaOH = 0.10 M = 0.10 mol/L

The reaction balanced reaction between HCl and NaOH is

  
HCI + NaOH + NaCl + H2O
............(1)

Now calculate moles of HCl and NaOH

Formula: -

  
Moles = Molarity x Volume(L)

Therefore,

Moles of HCl =
0.15(mol/L) 0.0145(L) = 0.002175(mol)

and

Moles of NaOH =   
0.10(mol/L) x 0.0053(L) = 0.00053(mol)

Now,

We construct an ICE table for reaction (1)  

	HCl	+	NaOH	$\large \rightarrow$	NaCl	+	H2O
initial	0.002175		0.00053		0		-
change	- 0.00053		- 0.00053		+ 0.00053		-
equilibrium	0.001645		0		0.00053		-

Total volume of Mixture = (14.5 ml + 5.3 ml) = 19.8 ml = 0.0198 L

From ICE table we have,

the moles of HCl left = 0.001645 mol

Since, HCl is a strong acid. It dissociates completely.

Thus, moles of HCl = moles of H+ ion = 0.001645 mol

Now, calculate the concentration of H+ ion in the mixture

Thus,

  
= MolesOfH+Ion VolumeOf Mixture

i.e.

   
0.001645(mol) (H+) = 2 = 0.0831(mol/L) = 0.0831(M) 0.0198(L)

We known that,

  
pH = -log[H+

therefore,

pH = -log(0.0831) = -(-1.08) = 1.08

Therefore,

The pH of a solution made by mixing 14.5 ml of 0.15 M HCl with 5.3 ml of 0.10 M NaOH is 1.08.