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:
[tex]n_1M_1V_1=n_2M_2V_2[/tex]
where,
[tex]n_1,M_1\text{ and }V_1[/tex] are the n-factor, molarity and volume of acid which is [tex]HCl[/tex]
[tex]n_2,M_2\text{ and }V_2[/tex] are the n-factor, molarity and volume of base which is KOH
We are given:
[tex]n_1=1\\M_1=0.50M\\V_1=550.0mL\\n_2=1\\M_2=0.160M\\V_2=?[/tex]
Putting values in above equation, we get:
[tex]1\times 0.150\times 550.0=1\times 0.160\times V_2\\\\V_2=516mL=0.516L[/tex] (1L=1000ml)
Thus volume in liters to three significant figures is 0.516 L
