Can you please solve question 53? Thank you! hat volume (mlL) of 748 x 10 M penc
ID: 552087 • Letter: C
Question
Can you please solve question 53?Thank you! hat volume (mlL) of 748 x 10 M penchlori acid can he neutralized with 115 ml. of 0.244 M sodium hydroxide? a. 125 b. 8.60 c. 188 d. 750 375 V,375 m A 52, what volume (mL) of 7.48 × 102 M phosphoric acid can be neutralized with 115 mL of 0.244 M sodium hydroxide? a 125 b. 375 c. 750 d. 188 e.75.0 E ? 53. what is the molarity of a NaOH solution if 28.2 mL of a 0.355 M H2SO4 solution is required to neutralize a 250-mL sample of the NaOH solution? o. 3 ss H ( u, o8d ): g . ul muls a. 0.801 b. 0.315 c.0.629 d. 125 (.0.400 54· Lead ions can be precipitated from aqueous solutions by the addition of aqueous iodide: Pt2+ (aq) + 21, (aq) PbI2 (s) Lead iodide is virtually insoluble in water so that the reaction appears to go to completion. How many milliliters of 3.550 M HICaq) must be added to a solution containing 0.700 mol of PbNo3)2 (aq) to completely precipitate the lead? a. 2.54 x 103 (b, 394 c. 197 d. 0.197 e. 0.394 M: 3.SSo 374 mL
Explanation / Answer
Ans. Balanced Reaction: H2SO4 + 2 NaOH ---------> Na2SO4 + 2 H2O
Stoichiometry: 2 mole NaOH is neutralized by 1 mol H2SO4.
# Moles of H2SO4 consumed = (Molarity x Volume of solution in liters) of H2SO4
= 0.355 M x 0.0282 L ; [1 M = 1 mol/ L]
= (0.355 mol / L) x 0.0282 L
= 0.010011 mol
# According to the stoichiometry of balanced reaction, 1 mol H2SO4 neutralizes 2 moles NaOH.
So,
Moles of NaOH in sample = 2 x Moles of H2SO4 consumed
= 2 x 0.010011 mol
= 0.020022 mol
Given, volume of NaOH solution = 25.0 mL = 0.025 L
Now,
Molarity of NaOH = moles of NaOH / Volume of solution in liters
= 0.020022 mol / 0.025 L
= 0.80088 mol/ L
= 0.801 M
So, correct option is- a. 0.801 M