10 mL of 2(M) NaOH solution is added to 200 mL of 0.5 (M) of NaOH solution. What is the final concentration ?

Question

10 mL of 2(M) NaOH solution is added to 200 mL of 0.5 (M) of NaOH solution. What is the final concentration ? (A) 0.57 (M) (B) 5.7 (M) (C) 11.4 (M) (D) 1.14 (M)

Tardigrade Admin · Accepted Answer

From molarity equation M1 V1+M2 V2 =MV(total) 2×(10/1000)+0.5×(200/1000)=M ×(210/1000) 120=M ×210 M=(120/210)=0.57 M

Q. $10 m L$ of $2 (M)$ $N a O H$ solution is added to $200 m L$ of $0.5 (M)$ of $N a O H$ solution. What is the final concentration ?

0.57 (M)

5.7 (M)

11.4 (M)

1.14 (M)

From molarity equation
$M_{1} V_{1} + M_{2} V_{2} = M V_{(t o t a l)}$
$2 \times \frac{10}{1000} + 0.5 \times \frac{200}{1000} = M \times \frac{210}{1000}$
$120 = M \times 210$
$M = \frac{120}{210} = 0.57 M$

Q. 10mL of 2(M) NaOH solution is added to 200mL of 0.5(M) of NaOH solution. What is the final concentration ?