A student finds that 22.01mL of 0.1038 M NaOH are required to completely neutral

Question

A student finds that 22.01mL of 0.1038 M NaOH are required to completely neutralize 30.00mL of HCl. A. Write a balanced chemical equation for this neutralization B. Determine the concentration(M) of the acid A student finds that 22.01mL of 0.1038 M NaOH are required to completely neutralize 30.00mL of HCl. A. Write a balanced chemical equation for this neutralization B. Determine the concentration(M) of the acid A student finds that 22.01mL of 0.1038 M NaOH are required to completely neutralize 30.00mL of HCl. A. Write a balanced chemical equation for this neutralization B. Determine the concentration(M) of the acid A. Write a balanced chemical equation for this neutralization B. Determine the concentration(M) of the acid

Explanation / Answer

Answer – Given, [NaOH] = 0.1038 M, volume = 22.0 mL

Volume of HCl = 30.0 mL

A)Balanced chemical equation for this neutralization -

HCl + NOH ----> NaCl + H2O

B) Concentration (M) of the acid

First we need to calculate the moles o NaOH

Moles of NaOH = 0.1038 M * 0.022 L

                           = 0.00228 moles

From the balanced reaction

1 moles of NaOH = 1 moles of HCl

So, 0.00228 moles of NaOH = ?

= 0.00228 moles of HCl

Now we know

Concentration (M) = moles / volume (L)

So, Concentration (M) of HCl = 0.00228 moles / 0.030 L

                                                 = 0.0761 M

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