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(Solved): please solve for a guaranteed thumbs up You begin with a flask containing 25.0mL of 0.400 ...


You begin with a flask containing \( 25.0 \mathrm{~mL} \) of \( 0.400 \mathrm{M} \mathrm{HNO}_{3}(\mathrm{aq}) \) and titrate

3. Calculate the \( \mathrm{pH} \) of the solution after \( 14.5 \mathrm{~mL} \) of the \( 0.800 \mathrm{M} \mathrm{NaOH} \)

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You begin with a flask containing of and titrate by adding (aq) 3. Calculate the of the solution after of the solution has been added. (2 points)


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25 ml of 0.400 M HNO3 =0.400×25=10 millimolExplanation:here strong acid and strong base titration. so simple use formula eq-1 and eq-2
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