Calculate the molarity of a NaOH solution if 27.3 mL of it reactions reacts with

Question

Calculate the molarity of a NaOH solution if 27.3 mL of it reactions reacts with 0.4084 g of KHP (Potassium hydrogen phthalate KC_8H_4O_4H, one protic hydrogen available, molar mass: 204.2 g/mol). Calculate the molarity of a sulfuric acid solution if 23.2 mL of it reacts with 0.212 g of Na_2CO_3. A volume of 40.0 mL of iron (II) sulfate is oxidized to iron (III) by 200 mL of 0.100 M potassium dichromate solution. What is the concentration of the iron (II) sulfate solution? 6Fe^3+ + Cr_2O_2^2- + 14 H^- rightarrow 6 Fe^3+ + 2 Cr^3+ + 7 H_2O Write the correct net ionic equation for the following reactions KOH + H_2SO_4 Ba(OH) + H_2CO_3 HF + Ca(OH)_2 H_3PO_4 + Mg (OH)_2 HCl + NH_3 HCIO + LiOH HClO_4 + Al(OH)_3 NaOH + Al(OH)_3 CH_3COOH (acetic acid) + Mg (OH)_2 KOH + Zm(OH)_2

Explanation / Answer

Q1

M of NaOH given

V= 27.3 mL of base

m = 0.4084 g of KHP

mol KHP = mass/MW = 0.4084/204.2 = 0.002 mol of KHP

ratio is 1:1

so

1 mol of base = 1 mol of acid

0.002 mol of acid = 0.002 mol of base

then

[NaOH] = mol/V = 0.002/(27.3*10^-3) = 0.07326 M of NaOH

b)

M of H2SO4 for

V = 23.2 mL of acid when

mol of Na2CO3= mass/MW = 0.212/105.9888 = 0.002

M= mol/V = 0.002/(23.2*10^-3) = 0.0862 M of Na2CO3

note that ratio is

2:2

Na2CO3 + H2SO4 = Na2SO4 + CO2 + H2O

so

0.0862 mol of Na2CO# = 0.0862mol of H2SO4

[H2SO4] = 0.0862 M

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