Question

What volume (mL) of 1.50 × 10-2 M hydrochloric acid can be neutralized with 115 mL of 0.244 M sodium hydroxide?

Answer 1

Vacid = 1870.66 mL

Step-by-step explanation:

neutralization point:

• (V*C)acid = (V*C)base

⇒ Vacid = ((0.115 L)*(0.244 M))/(1.50 E-2 M)

⇒ Vacid = 1.8706 L

⇒ Vacid = 1870.66 mL

Answer 2

Answer: 1870. 67mL

Step-by-step explanation:

NaOH + HCl —> NaCl + H2O

nA = 1, Ca = 0.015M, Va =?

nB = 1 Cb = 0.244 M, Vb = 115 mL

CaVa / CbVb = nA/nB

(0.015xVa)/(0.244x115) = 1/1

Va = (0.244x115) /0.015 = 1870. 67mL