If it took 30.0 mL of 0.50 M NaOH to neutralize 15.0 mL of HNO3, what is the concentration of HNO3? Explain and show how titration is used to find the amount of NaOH it took to neutralize the HNO3.

Question

simplymarie_x · Answer

@aaronq @Cuanchi

photon336 · Answer

$$Molarity_{HNO_{3}}*Volume_{HNO_{3}} = Molarity_{NaOH}*Volume_{NaOH}$$

simplymarie_x · Answer

NaOH + HNO3 ==> NaNO3 + H2O balanced equation 
moles of NaOH used = 0.030 L x 0.50 mol/L = 0.0150 moles NaOH used 
moles of HNO3 neutralized = 0.0150 moles HNO3 because 1 mole NaOH = 1 mole HNO3 (see balanced eq) 
concentration of HNO3 = moles/L = 0.0150 moles/0.0150 L = 1.0 M

Is that right?

photon336 · Answer

Yep

simplymarie_x · Answer

Does that also explain how titration is used? @Photon336

photon336 · Answer

If I remember, in a titration you add a certain amount of titrant, until you reach the end point, the point where the number of moles of acid and base are equal.

simplymarie_x · Answer

Okay thank you ^~^

photon336 · Answer

no problem