Question

297 Experiment 24 10. Common ion effect a. Concentration of KHT final solution (50 mL) b. Volume of solution c. Mass of KCI d. Concentration of KCI 5.0x10 mL 271 381 f. Total [H [KHned g. Saturation temperature (C) h. Saturation temperature (K) 516 332.7S 0428 j. Why did the temperature have to be increased to dissolve the KHT when the KCI was added? B. K, by titration. 10.1:C 1. Temperature of water bath 2. Volume of KHT aliquot 3. Molarity of NaOH solution 4. Final buret reading 5. Initial buret reading 6. Volume of NaOH 7. Moles of NaOH 8. Moles of KHT 9. [K.] = [HT] lloml -3
i need help with number 8 and 9 please!

Answer 1

Solution :-

KHT + NaOH --- > KNaT + H2O

Mole ratio of the KHT and NaOH is 1 : 1

Therefore the moles of KHT are same as moles of NaOH

#8) So the moles of KHT = 0.0011 mol

#9) [K^+]=[HT^-] = moles / volume

                         = 0.0011 mol / 0.010 L

                        = 0.110 M