Question
Asked Oct 7, 2019

If 250.0 mL of 9.84 x 10-4  M HNO3 is mixed with 150.0 mL of 1.18 x 10-5 M HBr, what is the resulting pH?

 
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Step 1

HNO3 and HBr are strong acids and hence dissociates completely. The number of moles of H+ ions formed by the dissociation of HNO3 and HBr can be calculated as follows.

Also, 1 mL = 0.001 L

No. of moles of HNO;- Molarity x Volume
= (9.84x10 M(0.25 L) 2.46 x 10 mol
H NO
On dissociation: HNO
Hence, 2.46x10
mol H* ions are formed on dissociation of HNO;
3
No. of moles of HBr = Molarity x Volume
= (1.18x10 M) (0.15 L) = 1.77x 10
mol
>H Br
On dissociation: HBr
Hence, 1.77x10 mol H* ions are formed on dissociation of HBr
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No. of moles of HNO;- Molarity x Volume = (9.84x10 M(0.25 L) 2.46 x 10 mol H NO On dissociation: HNO Hence, 2.46x10 mol H* ions are formed on dissociation of HNO; 3 No. of moles of HBr = Molarity x Volume = (1.18x10 M) (0.15 L) = 1.77x 10 mol >H Br On dissociation: HBr Hence, 1.77x10 mol H* ions are formed on dissociation of HBr

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Step 2

Concentration of H+ ions of the resulting solution formed by the mixing...

= (2.46x 10 mol)+(1.77 x10 mol)
Total no. of moles of H ions
2.478 x 10
mol
Total volume = 250 mL 150 mL = 400 mL = 0.40 L
No. of moles
Concentration of H ions
Volume(in L
2.478x 10 mol
= 6.195x 105 M
14
0.40 L
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= (2.46x 10 mol)+(1.77 x10 mol) Total no. of moles of H ions 2.478 x 10 mol Total volume = 250 mL 150 mL = 400 mL = 0.40 L No. of moles Concentration of H ions Volume(in L 2.478x 10 mol = 6.195x 105 M 14 0.40 L

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