What quantity of NaOH(s)must be added to 2.00 L of 0.434 M HCl to achieve a pH of 13.00? (Assume no volume change. )

Question

Quizplus · Accepted Answer

To achieve a pH of 13, we need a basic solution with a pOH of 1 (14 - pH = pOH). From the pOH, we can calculate the hydroxide ion concentration to be 1.00 x 10^-1 M. To neutralize this amount of OH-, we need an equal amount of H+ ions from the HCl. This amounts to 1.00 x 10^-1 mol of HCl.

Using the formula M1V1 = M2V2, we can calculate the volume of NaOH needed to neutralize this amount of HCl:

(0.434 mol/L) (2.00 L) = (x mol/L) (V2)
V2 = 0.434 mol / x

Since we need 1.00 x 10^-1 mol of NaOH to neutralize the HCl, we can set up another equation using the molarity and volume of NaOH:

x mol/L * V2 = 1.00 x 10^-1 mol
x = 1.07 mol

Therefore, the quantity of NaOH that must be added is 1.07 mol.

What Quantity of NaOH(s)must Be Added to 2 ×\times× 10-13 Mol E)none of These

What Quantity of NaOH(s)must Be Added to 2 $\times$ 10^-¹³ Mol
E)none of These