Twenty-five milliliters of 0.10 M HCl is titrated with 0.10 M NaOH. What is the pH after 15 ml of NaOH has been added

The correct answer is 1.60.

Explanation:

Based on the given question, 25 ml of 0.10 M HCl is titrated with 0.10 M NaOH. Now moles of HCl can be determined by using the formula,  

Moles = volume * concentration of HCl

= 25/1000*0.10 = 0.0025 moles

Similarly the moles of NaOH added will be determined by using the formula,  

Moles of NaOH added = volume * concentration of NaOH

= 15/1000 * 0.10 = 0.0015 moles

The reaction taking place in the given case is,  

HCl + NaOH = NaCl + H2O

Now the moles of excess H+ = moles of excess HCl

= 0.0025 - 0.0015 = 0.001 moles

Based on the given question, the sum of the volume of the solution is 25+15 = 40 ml or 0.040 L

[H+] = moles of H+/total volume

= 0.001 / 0.040 = 0.025 M

pH = -log[H+]

= -log[0.025]  

= 1.60