Question

2. Calculate the mass of solid sodium hydroxide needed to prepare 300 mL of 0.1 M sodium hydroxide. Note: Be sure to include this mass in the procedure for Part C. 3. A 0.6250 g sample of KHP (204.2 g/mol), dissolved in 50 mL water, required 28.50 mL of sodium hydroxide to reach the end point. What is the concentration of the sodium hydroxide solution? 4. A 25.00 mL sample of hydrochloric acid was titrated using the same sodium hydroxide solution from Question #4. Given that the initial and final buret readings were 2.90 mL and 26.64 mL, respectively, what is the concentration of the hydrochloric acid? 5. A student dissolved 2.5123 g sample of soda ash into 100.00 mL of deionized water and titrated 25.00 mL of this solution with 0.1020 M HCl. It took 33.20 mL of acid to titrate the solution to the second equivalence point. a. Calculate the concentration of the sodium carbonate solution (Calculation #4, pg. 9).

Answer 1

kindly post different questions separately. please rate

Volume of NaOH solution - 300 mL = 0.3L (1000 mL = 1L) 0.IM concentration of NaOH = moles of NaOH needed = concentration (M) x volume (ll - 0.1M X 0,3L = NOOH = 0.03 mol 40 g/mol vel molar mass of mass of NaOH needed = molar mass X moles - 40g/mol X 0.03 mol 1.2 g of wooh Answer