A 35.00 mL sample of NaOH is titrated to an endpoint by 14.76 mL 0.4122M HBr. What's the Molarity of theNaOH?

Answer 1

HBr is a strong acid

HBr>> H+ + Br-

Moles H+ = 14.76 x 0.4122 / 1000 = 0.00608

At the endpoint moles H+ = moles OH-

[ OH- ] = 0.00608 / 0.035 L = 0.174 M

Answer 2

Firstly, write a balanced equation for the reaction between HBr and NaOH.

HBr + NaOH --> H2O + NaBr

0.4122M = 0.4122 moles per litre

Thus, the number of moles of HBr = (14.76/1000) x 0.4122 = 0.006084072

We can also deduce that the number of moles of NaOH = 0.006084072, due to the ratio being 1:1

35.00 mL sample of NaOH gives 0.006084072 moles
1000ml sample of NaOH gives 0.17383 moles = 0.174M

Answer 3

Na OH +HBr ---> NaBr +H2O

remeber this formula if you have titration by 2 reactants with the same valence, (here the valence is 1)

M1V1 =M2V2 let NaOH be 1 and HBr be 2

so 35* m(NaOH) = 0.4122*14.76

so m(NaOH) = 0.4122*14.76/35= 0.1738M

Answer 4

Your answers are too long.

M NaOH
= (mLM HBr * ratio) / mL NaOH
= [14.76 mL * 0.4122M * (1 NaOH/ 1HBr)] / 35 mL
=0.1738