To determine the pH of the resulting solution, we need to first calculate the moles of HNO3 and NaOH. The equation for the reaction between HNO3 and NaOH is:

HNO3 + NaOH → NaNO3 + H2O

Using the formula: Moles = Molarity x Volume (in liters), we can calculate the moles of HNO3 and NaOH:

Moles of HNO3 = 0.2 M x 0.04 L = 0.008 mol
Moles of NaOH = 0.3 M x 0.06 L = 0.018 mol


The limiting reactant is the reactant that is completely consumed in the reaction. To determine the limiting reactant, we compare the moles of HNO3 and NaOH.

From the calculations above, we can see that the HNO3 has fewer moles compared to NaOH. Therefore, HNO3 is the limiting reactant.


Since HNO3 is the limiting reactant, NaOH will be present in excess. To calculate the moles of excess NaOH, we subtract the moles of HNO3 from the moles of NaOH:

Moles of excess NaOH = 0.018 mol - 0.008 mol = 0.01 mol


To calculate the concentration of the resulting solution, we add the volumes of HNO3 and NaOH and divide the total moles by the total volume:

Total volume = 40 mL + 60 mL = 100 mL = 0.1 L
Total moles = Moles of HNO3 + Moles of NaOH = 0.008 mol + 0.018 mol = 0.026 mol

Concentration of resulting solution = Total moles / Total volume = 0.026 mol / 0.1 L = 0.26 M


To calculate the pH of the resulting solution, we need to determine the concentration of H+ ions. In this case, the resulting solution is a neutralization reaction between a strong acid (HNO3) and a strong base (NaOH), producing water and a salt (NaNO3). Therefore, the concentration of H+ ions is equal to the concentration of OH- ions, which can be calculated using the concentration of the resulting solution:

[H+] = [OH-] = 0.26 M

To calculate the pH, we use the formula: pH = -log[H+]

pH = -log(0.26) = 0.59

Therefore, the pH of the resulting solution is approximately 0.59.

N=nbvb-nava÷va+vb
N=12-8÷100
N=4÷100
pH =-log(H) +
pH=-log4÷100
PH=-0.6020+2
PH=1.3980