A 25.0 mL sample of 0.105 M $HCl$ was titrated with 315 mL of $NaOH$ . What is the concentration of the $NaOH$ ?

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Answer 1 · 2017-07-10T03:28:57

$0.00833M$

We're asked to find the molar concentration of the $"NaCl"$ solution given some titration data.

Let's first write the chemical equation for this reaction:

$"NaOH"(aq) + "HCl" (aq) rarr "NaCl" (aq) + "H"_2"O" (l)$

Using the molarity equation, we can find the number of moles of $"HCl"$ that reacted:

$"molarity" = "mol solute"/"L soln"$

$"mol solute" = ("molarity")("L soln")$

$"mol HCl" = (0.105"mol"/(cancel("L")))(0.0250cancel("L")) = 0.00263$ $"mol HCl"$

(volume converted to liters)

Now, using the coefficients of the chemical reaction, we can determine the number of moles of $"NaOH"$ that reacted:

$0.00263cancel("mol HCl")((1color(white)(l)"mol NaOH")/(1cancel("mol HCl"))) = 0.00263$ $"mol NaOH"$

Lastly, we'll use the molarity equation (using given volume of $"NaOH soln"$ ) again to determine the molarity of the sodium hydroxide solution:

$"molarity" = "mol solute"/"L soln"$

$M_ "NaOH" = (0.00263color(white)(l)"mol")/(0.315color(white)(l)"L") = color(red)(0.00833M$

(volume converted to liters)

A 25.0 mL sample of 0.105 M HClHCl was titrated with 315 mL of NaOHNaOH. What is the concentration of the NaOHNaOH?