During a titration, 60.0 mL of 0.010 M NaOH is added to 35.0 mL of 0.020 M HCl solution. What is the pH of the resulting solution?

Question

Quizplus · Accepted Answer

First, we need to calculate the number of moles of NaOH and HCl:

moles of NaOH = volume (in L) x concentration 
= 0.060 L x 0.010 mol/L 
= 0.0006 mol

moles of HCl = volume (in L) x concentration 
= 0.035 L x 0.020 mol/L 
= 0.0007 mol

Since the reaction between NaOH and HCl is a neutralization reaction, we can use the mole ratio of NaOH to HCl to determine the limiting reactant. The balanced chemical equation for the reaction is:

NaOH + HCl → NaCl + H2O

The mole ratio of NaOH to HCl is 1:1, so in this case, HCl is the limiting reactant since there are more moles of it.

Next, we need to determine the amount of excess HCl that remains after all the NaOH has reacted. The moles of HCl that react with NaOH is also 0.0006 mol:

moles of HCl that react with NaOH = 0.0006 mol

The remaining moles of HCl is:

moles of HCl remaining = 0.0007 mol – 0.0006 mol 
= 0.0001 mol

Now, we can calculate the concentration of HCl in the resulting solution:

concentration of HCl = moles/volume 
= 0.0001 mol / (0.060 L + 0.035 L) 
= 0.00087 mol/L

To determine the pH, we need to use the following equation:

pH = -log[H+], where [H+] is the concentration of hydrogen ions

In this case, the hydrogen ions come from the dissociation of HCl:

HCl → H+ + Cl-

The concentration of hydrogen ions is equal to the concentration of HCl since HCl dissociates completely in water:

[H+] = 0.00087 mol/L

pH = -log(0.00087) = 2.98

Therefore, the pH of the resulting solution is 2.98, and the correct answer is C.

During a Titration, 60