What volume of a 1.0 M HCI is required to completely neutralize 25.0 ml of a 1.0 M KOH?

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What volume of a 1.0 M HCI is required to completely neutralize 25.0 ml of a 1.0 M KOH?

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Answer 1 · 2017-04-07T05:17:30

Why, an equal volume, i.e. a $25.0 \cdot m L$ $25.0*mL$ volume............

Explanation:

We interrogate the reaction:

$K O H (a q) + H C l (a q) \to K C l (a q) + H_{2} O (l)$ $KOH(aq) + HCl(aq) rarr KCl(aq) + H_2O(l)$

When we quote solution concentration, we refer to the quotient:

$Concentration$ $"Concentration"$ $=$ $=$ $\frac{Moles of solute}{Volume of solution}$ $"Moles of solute"/"Volume of solution"$ , i.e.

$C = \frac{n}{V}$ $C=n/V$ , and thus $concentration$ $"concentration"$ has units of $m o l \cdot L^{- 1}$ $mol*L^-1$

And you will habitually use this quotient to calculate to calculate solution concentrations, moles and mass of solute, and volume of solution needed.

It should not take a great deal of insight to appreciate that the given volume of $N a O H (a q)$ $NaOH(aq)$ solution contains THE SAME AMOUNT OF SOLUTE that is in the given volume of $H C l (a q)$ $HCl(aq)$ .

i.e. $moles of KCl$ $"moles of KCl"$ $=$ $=$

$concentration \times volume = 1.0 \cdot m o l \cdot L^{- 1} \times 25.0 \times 10^{- 3} \cdot L$ $"concentration"xx"volume"=1.0*mol*L^-1xx25.0xx10^-3*L$

$= 25 \times 10^{- 3} \cdot m o l$ $=25xx10^-3*mol$ .

What is the final concentration of $K C l (a q)$ $KCl(aq)$ ? Why is it different from the starting concentrations of $[K O H]$ $[KOH]$ , and $[H C l]$ $[HCl]$ ?

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What volume of a 1.0 M HCI is required to completely neutralize 25.0 ml of a 1.0 M KOH?

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