Calculate the molarity (M) of KOH solution which required 15.20 mL of KOH to rea

Question

Calculate the molarity (M) of KOH solution which required 15.20 mL of KOH to react completely with 0.651 g of KHP(Molar mas of KHP is 204.2g). This problem is similar to what is done in the lab report and show in your lab manual. Shows all steps (give step number 1, 2, 3 etc. for each step of calculation) by using conversion factors, units and finish with c.v. and F.v. KOH(aq) + KHP(aq) rightarrow KKP(aq) + H_2O(l) In an experiment to calculate heat of neutralization reaction using a calorimeter uVmg a coffee cup, a group of students added 1.969 g of solid NaOH to 100.0 ml (100.0 g) of 0.50 M HCl solution. The initial temperature of the solution was 22.5 degree C and final temperature measured was 33.9 degree C. Calculate the heat of neutralization (q) in this reaction. (use SH of water as 4.18 J/g- degree C). Show all steps of calculations. Give final value of q in J Finish with KJ. Show Table values Mass of solution Delta t degree C = SH = Show formula of q and substitute Catenation step with units

Explanation / Answer

Given

Mass = 0.651 g

Molar mass = 204.2 g/mol

No. of moles = Mass / Molar mass = 0.651 g / 204.2 g/mol = 3.19 * 10-3 mol

No. of moles of KHP used is equal to no.of moles of KOH reacting

No. of moles of KOH = 3.19 * 10-3 mol

Volume = 15.2 ml = 0.0152 L

No. of moles = Volume * Moalrity

3.19 * 10-3 mol = 0.0152 L * M

M = 0.21 mol/L

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