Question

A 40.00 cm³ of 0.100 mol dm³ HCL solution is titrated with 0.100 mol dm³ of NaOH Calculate the pH after adding 10.00 cm³ of NaOH to a 40.00 cm³ of 0.100 mol dm³ HCl solution.

a) 1.00
b) 1.30
c) 2.00
d) 2.30

Answer 1

The pH value after adding 10.00 cm³ of 0.100 mol dm³ NaOH to a 40.00 cm³ of 0.100 mol dm³ HCl solution is approximately 1.22, which corresponds to answer option (b) 1.30.

Step-by-step explanation:

When you add 10.00 cm³ of 0.100 mol dm³ NaOH to a 40.00 cm³ of 0.100 mol dm³ HCl solution, a neutralization reaction occurs where HCl and NaOH react in a 1:1 molar ratio to form NaCl and water. Since both solutions have the same molarity, the amount of NaOH added is not enough to completely neutralize the HCl. To calculate the remaining concentration of HCl, you subtract the moles of NaOH from the moles of HCl initially present:

• Moles of HCl initially = 40.00 cm³ × 0.100 mol/dm³ = 0.00400 mol
• Moles of NaOH added = 10.00 cm³ × 0.100 mol/dm³ = 0.00100 mol
• Moles of HCl remaining = 0.00400 mol - 0.00100 mol = 0.00300 mol

The total volume after addition is 50.00 cm³. Thus, the concentration of HCl remaining is 0.00300 mol / 0.05000 dm³ = 0.0600 M. To find the pH:

pH = -log[H+] = -log[0.0600] ≈ 1.22

Since the exact pH value of 1.22 is not one of the options given, we would round it to the nearest tenth, considering the number of significant figures provided in the options. Thus, the closest answer would be b) 1.30.