Question

QUESTION 4 What is the hydronium-ion concentration of a solution formed by combining 400. mL of 0.21 M HNO 3 with 600. mL of 0.11 M NaOH at 25 C7 OA. 0.082 M OB. 6.7x 10-13 M OC.0.018 M O D.0.11 M O E. 0.21 M

Answer 1

The concentration of HNO₃ = 0.21 M

The volume of HNO₃ = 400. mL x( 1L/1000 mL) = 0.400 L

The concentration of NaOH= 0.11 M

The volume of NaOH = 600. mL x (1 L/1000 mL) = 0.600 L

Determine the number of moles of HNO₃ and NaOH from the given data as follows:

For HNO₃:

= 0.21 M x 0.400 L

= 0.084 mol HNO₃

Similarly, for NaOH,

= 0.11 M x 0.600 L

= 0.066 mol NaOH

The balanced reaction between HNO₃ and NaOH is as follows:

HNO₃(aq) + NaOH(aq) $\rightarrow$ NaNO₃(aq) + H₂O(l)

Drawing an ICE chart as follows:

	HNO3(aq)	NaOH(aq)	NaNO3(aq)
I(moles)	0.084	0.066	0
C(moles)	-0.066	-0.066	+0.066
E(moles)	0.018	0	0.066

Thus,

Number of moles of HNO₃ in the solution is 0.018 mol.

Total volume of solution = 0.4 L + 0.6 L = 1.0 L

[HNO₃] = 0.018 mol / 1 L

[HNO₃] = 0.018 M

Since, HNO₃ is a monoprotic strong acid, the hydronium ion concentration of HNO₃ is nothing but the concentration of HNO₃.

HNO₃ dissociates as follows:

HNO₃(aq) + H₂O(l) $\rightarrow$ H₃O⁺(aq) + NO₃^-(aq)

Thus,

[H₃O⁺] = 0.018 M