25 cm3 of NaOH was titrated with 0.050 mol dm-3 HCl. NaOH + HCl --> NaCl + H2O. 21.5 cm3 HCl neutralised 25 cm3 NaOH. Concentration of NaOH in mol dm-3?

21.5 cm3 / 1000 = 0.0215 dm3 of HCl. n = cv so (0.050) x (0.0215) = 0.001075 mols of HCl. NaOH + HCl --> NaCl + H2O so there is a 1:1 ratio of reactants to products. Therefore there are 0.001075 mols of NaOH. n = cv so n /v = c. (0.001075) / (0.025) = 0.043 mol dm-3 is the concentration of NaOH

Answered by Arran S. Chemistry tutor

2337 Views

See similar Chemistry GCSE tutors

Related Chemistry GCSE answers

All answers ▸

Sodium hydroxide sol'n solution is poured into a beaker of hydrochloric acid which contains a thermometer showing room temperature. a) What is the name of the product produced? b) What type of reaction does the thermometer's temperature rise prove?


What is made when an acid is added to an alkali?


Define an isotope.


Explain the significance of atomic number and mass number in determining atomic properties. Which of these changes in isotopes?


We're here to help

contact us iconContact usWhatsapp logoMessage us on Whatsapptelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo
Cookie Preferences