Question

A 25.0-mL sample of H2SO4 is neutralized with NaOH. What is the concentration of the H2SO4
if 35.0 mL of 0.150 M NaOH are required to
completely neutralize the acid?
Can someone please help me with this question and explain it as well thank you?

Answer 1

Theres a shortcut you can take for this question, but i think you should work through it first, so you understand why it works.

Treat this like any stoichiometry problem. First, write a balanced equation for the reaction and find the number of moles of both participants, using:

\(Molarity=\dfrac{moles}{L}\)

ps. you need to convert mL to L; 1 L = 1000 mL

Answer 2

once you do that, use the stoichiometric coefficients to find moles of acid that were neutralized, set up a ratio using the species of interest, like so:

e.g. for a general reaction:

\(\color{red}{a}A + \color{blue}{b}B\) \(\rightleftharpoons\) \( \color{green}{c}C\)

where upper case are the species (A,B,C), and lower case (a,b,c) are the coefficients ,

\(\dfrac{n_A}{\color{red}{a}}=\dfrac{n_B}{\color{blue}{b}}=\dfrac{n_C}{\color{green}{c}}\)

From here you can isolate what you need.
For example: if you have 2 moles of B, how many moles of C can you produce?

solve algebraically: \(\dfrac{2}{\color{blue}{b}}=\dfrac{n_C}{\color{green}{c}}\rightarrow n_C=\dfrac{2*\color{green}{c}}{\color{blue}{b}}\)
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After, use the moles and the volume (given in the question) to find the concentration (molarity) of the acid using the formula i first posted.