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Consider the neutralization reaction2 HNO3(aq) + Ba(OH)â‚‚ (aq) -> 2 Hâ‚‚O(l) + Ba(NO3)â‚‚(aq)A 0.125 L sample of an unknown HNO, solution required 42.7 mL of 0.100 M Ba(OH)â‚‚ for complete neutralization. What isthe concentration of the HNO, solution?

Respuesta :

Answer:

C = 0.068 M

Explanation:

We are given: volume of HNO = 0.125 L

: volume of Ba(OH)2 = 42.7 mL

: concentration of Ba(OH)2 = 0.100 M

We first determine the number of moles of Ba(OH)2:

n = CV

Where: n is number of moles, C is concentration, and V is volume

n = CV

= 0.0427 * 0.100

= 0.00427 mol

From the balanced equation:

n(HNO3) = 2 * n(Ba(OH)2)

= 2 * 0.00427

= 0.00854

For HNO3:

C = n/V

= 0.00854/0.125

= 0.068 M

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