Question

What volume of a 0.160 MM solution of KOHKOH must be added to 550.0 mLmL of the acidic solution to completely neutralize all of the acid? Express the volume in liters to three significant figures.

Answer 1

The given question is incomplete. The complete question is :

What volume of 0.160 M solution of KOH must be added to 550.0 mL of the acidic solution to completely neutralize all of the 0.150 M hydrochloric acid?

Answer: Volume in liters to three significant figures is 0.516 L

Explanation:

To calculate the volume of acid, we use the equation given by neutralization reaction:

$n_1M_1V_1=n_2M_2V_2$

where,

$n_1,M_1\text{ and }V_1$ are the n-factor, molarity and volume of acid which is $HCl$

$n_2,M_2\text{ and }V_2$ are the n-factor, molarity and volume of base which is KOH

We are given:

$n_1=1\\M_1=0.50M\\V_1=550.0mL\\n_2=1\\M_2=0.160M\\V_2=?$

Putting values in above equation, we get:

$1\times 0.150\times 550.0=1\times 0.160\times V_2\\\\V_2=516mL=0.516L$   (1L=1000ml)

Thus volume in liters to three significant figures is 0.516 L