A sample of 7.4 L of NH3 gas at 22 ∘C and 735 torr is bubbled into a 0.50-L solution of 0.35 M HCl. Assuming that all the NH3 dissolves and that the volume of the solution remains 0.50 L, calculate the pH of the resulting solution. Kb for Nh3=1.8x10-5

pH = 9.11

Explanation:

First, using gas law, we will determine the moles of NH3 bubbled. Then, with moles of HCl we can determine how many NH4⁺ are produced. As last, using H-H equation for bases we can find pOH and pH of the solution:

Moles NH3:

PV = nRT

P = pressure: 743 torr * (1atm / 760torr) = 0.9776atm

V = volume: 7.4L

n = Moles. Our incognite

R = Gas constant: 0.082atmL/molK

T is absolute temperature: 273.15K + 22°C = 295.15K

PV / RT = n

0.9776atm*7.4L / 0.082atmL*295.15K = 0.30 moles of NH3

Moles HCl:

0.50L * (0.35mol / L) = 0.175 moles of HCl

Based on the reaction:

NH3 + HCl → NH4⁺ + Cl⁻

All HCl added is producing NH₄⁺ = 0.175 moles NH₄⁺

Moles NH₃: 0.30moles NH3 - 0.175 moles = 0.125 moles NH₃

H-H equation for bases is:

pOH = pKb + log [BH+] / [B]

Where pKb = -log Kb = 4.74

[BH+] could be taken as moles of NH4+ = 0.175 mol

[B] are moles of NH3 = 0.125 mol

pOH = pKb + log [BH+] / [B]

pOH = 4.74 + log [0.175mol] / [0.125mol]

pOH = 4.89

pH = 14-pOH

answer: the enthalpy of the overall chemical reaction is +87.88 kj or +88 kj.

according to hess's law, the overall enthalpy of a reaction is the difference between the sum of enthalpy values of all the products and the sum of enthalpy values of all the reactants.

δh(reaction) = ∑nδh(products) - ∑nδh(reactants)

δh(reaction) = [δh +δh] - [δh]

δh(reaction) = [1(-287.02) + 1(0) ] - [1(-374.9)] kj = +87.88 kj

here, δh (pcl3) = -287.02 kj, δh(cl2) = 0, δh(pcl5) = -374.9 kj

therefore, enthalpy of the overall reaction is +87.88 kj or +88 kj.