Determine the concentration of a NaOH solution if 27.80mL of NaOH is required to neutralize 10.00mL of a 1.00 M H2SO4 solution. (Hint:Review monoprotic vs. diprotic acids.)

would I do (27.80)(10)?

Thank you for your help!!

Hannah, I can't believe you can't follow instructins but here is the problem worked IN DETAIL.

H2SO4 + NaOH ==> Na2SO4

moles H2SO4 in the problem = M x L = 1.00M x 0.01000 L = 0.01000 moles (10.00 mL = 0.01 moles).My instructions were moles H2SO4 = M x L.

Now we convert moles H2SO4 to moles NaOH using the coefficients in the balanced equation.
0.01000 moles H2SO4 x (2 moles NaOH/1 mole H2SO4) = 0.01000 moles x (2/1) = 0.02000 moles NaOH. My instructions were moles NaOH = 2xmoles H2SO4.

Then M NaOH = moles NaOH/L NaOH.
M NaOH = 0.02/0.02780 = 0.71942 M NaOH which rounds to 0.7194 M NaOH. My instructions were the same.

If you wish to work in millimoles (it keeps the fractions down) that will be
(10.00 x 1.00M x 2)/27.8 = 0.7194 M
If you still don't understand, please tell me EXACTLY what the trouble is.

2NaOH + H2SO4 ==> 2H2O + Na2SO4

moles H2SO4 = M x L = ?
moles NaOH = 2 x moles H2SO4 from the equation.
M NaOH = moles NaOH/L NaOH. You have moles NaOH and L NaOH, solve for M NaOH.

Acid- Base titration of a weak-acid. H2S04 is a weak acid. NaOH is a strong base.

pH = 14 + log(sqrt)[(M1*M2*Kw)/([M1+M2]*Ka)]

M1 = Concentration of Weak Acid.
M2 = Concentration of Strong Base.
Kw = dissociation constant for water.
Ka = dissociation constant for acid.

Note: M = moles/volume

Isolate for M1.

My response stands in spite of the wild formula posted. H2SO4 is a strong acid for the first H ionized and a weak (but still relatively strong) acid for the second H (k2 = about 0.012). However, with NaOH, the two H atoms cannot be titrated separately and both will titrate more or less at the same time. A slight change can be seen in a titration curve but there is no indicator that will detect the small change. The slope of the pH curve simply is not enough to use an indicator nor is it enough to use graphical methods.

SO would it be 27.80 / 10.00?

No. It isn't 27.8*10.00 nor is it 27.80/10.00? Follow my instructions.

ok i do not understand at all but thank you for your help.

Thank you!!

is the answer 0.72m?

To determine the concentration of the NaOH solution, you need to use the concept of stoichiometry and the balanced chemical equation for the reaction between NaOH and H2SO4.

First, let's write the balanced chemical equation for the reaction:

2NaOH + H2SO4 -> Na2SO4 + 2H2O

From the equation, we can see that 2 moles of NaOH react with 1 mole of H2SO4 to produce 2 moles of water (H2O) and 1 mole of Na2SO4.

Given that 27.80 mL of NaOH solution reacted with 10.00 mL of 1.00 M H2SO4 solution, we can determine the number of moles of H2SO4 used.

Using the equation: Molarity (M) = moles/volume (L), we can rearrange it to:

moles = Molarity x volume (L)

Molarity of H2SO4 = 1.00 M (given)
Volume of H2SO4 = 10.00 mL = 0.01000 L

moles of H2SO4 = 1.00 M x 0.01000 L = 0.01000 moles

Since the stoichiometry between NaOH and H2SO4 is 2:1 (2 moles of NaOH react with 1 mole of H2SO4), the number of moles of NaOH used can be determined:

moles of NaOH = 0.01000 moles x 2 = 0.02000 moles

Now, let's calculate the concentration of the NaOH solution:

Volume of NaOH solution used = 27.80 mL = 0.02780 L

Molarity of NaOH = moles of NaOH / volume of NaOH solution (L)

Molarity of NaOH = 0.02000 moles / 0.02780 L ≈ 0.719 M

Therefore, the concentration of the NaOH solution is approximately 0.719 M.