Question

estion 16 of 21 > Attempt 4 - A solution is made by mixing 500.0 mL of 0.03479 M Na, HAso, with 500.0 mL of 0.04083 M NaOH. Complete the mass balance expressions for the sodium and arsenate species in the final solution. [HASO-1+ H, AsO+HASO, + Aso - = .017395 [Nat] = .03781 terms of use contact us help bout us career privacy policy MacBook Pro

Answer 1

Sol :-

The mass balance equation is :

HAsO42-  + H3AsO4 + AsO4-3 + H2AsO4-

As,

Number of moles = Molarity x Volume of solution in L

So,

Total number of moles of Na⁺ = 0.03479 M x 0.500 L + 0.04083 M x 0.500 L

= 0.017395 mol + 0.020415 mol

= 0.03781 mol

Volume of the solution = 500 ml + 500 ml = 1000 ml = 1.0 L

So,

[Na⁺] = Moles of Na⁺ / Volume of solution in L

= 0.03781 mol / 1.0 L

= 0.03781 M

Hence, [Na+] = 0.03781 M

Similarly,

Total number of moles of HAsO₄^2-  = 0.03479 M x 0.500 L

= 0.017395 mol

Volume of the solution = 500 ml + 500 ml = 1000 ml = 1.0 L

So,

[HAsO₄^2- ] = Moles of HAsO₄^2-  / Volume of solution in L

= 0.017395 mol/ 1.0 L

= 0.017395 M

Hence, [HAsO42-] = 0.017395 M