The molar concentration of chloride ions in the resulting solution of 300 mL of 3.0 M NaCl and 200 mL of 4.0 M ` BaCl _ 2 ` will be

Options

A. `1.7 ` M
B. `1.8 ` M
C. ` 5.0 ` M
D. ` 3.5 ` M

Answer

C

Explanation

` underset (" 3.0 M") ( NaCl) hArr Na ^ + + underset("3.0 M")(Cl ^ - ) `
`underset ("4.0 M") (BaCl_2)hArr Ba^( 2 + ) + underset ( 2xx 4.0 M) ( 2Cl^-) `
` therefore ` Molar concentration of
` Cl^- = ( M _ 1 V _ 1 + M _ 2 V _ 2 ) /( V _ 1 + V _ 2 ) `
` = (3.0 xx 300 + 2 xx 4.0 xx 200 ) /( 300 + 200 ) `
` = ( 900 + 1600 )/( 500 ) `
` = 5.0 M `