Consider titration of NaOH solution versus 1.25 M oxalic acid solution. At the end point following burette readings were obtained. (i) 4.5 mL (ii) 4.5 mL (iii) 4.4 mL (iv) 4.4 mL (v) 4.4 mL If the volume of oxalic acid taken was 10.0 mL then the molarity of the NaOH solution is M. (Rounded-off to the nearest integer)

Question

Consider titration of NaOH solution versus 1.25 M oxalic acid solution. At the end point following burette readings were obtained. (i) 4.5 mL (ii) 4.5 mL (iii) 4.4 mL (iv) 4.4 mL (v) 4.4 mL If the volume of oxalic acid taken was 10.0 mL then the molarity of the NaOH solution is M. (Rounded-off to the nearest integer) (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

V NaOH =4.4 ml eq. of NaOH = eq. of H 2 C 2 O 4 or, M × 4.4 × 1=1.25 × 10 × 2 or, M =5.68 M ∴ Nearest integer answer is 6

VNaOH​=4.4ml eq. of NaOH= eq. of H2​C2​O4​ or, M×4.4×1=1.25×10×2 or, M=5.68M ∴ Nearest integer answer is 6

$V_{N a O H} = 4.4 m l$

eq. of $N a O H =$ eq. of $H_{2} C_{2} O_{4}$

or, $M \times 4.4 \times 1 = 1.25 \times 10 \times 2$

or, $M = 5.68 M$

$∴$ Nearest integer answer is 6