When 0.100 mole of ammonia, NH3 is dissolved in sufficient water to make 1.0 L of solution, the solution is found to have a hydroxide ion concentration of 1.34 × 10-3 M. Calculate Kb for ammonia.

Question

When 0.100 mole of ammonia, NH3 is dissolved in sufficient water to make 1.0 L of solution, the solution is found to have a hydroxide ion concentration of 1.34 × 10-3 M. Calculate Kb for ammonia. (A) 1.56 × 103 (B) 1.82 × 10-5 (C) 1.37 × 10-3 (D) 1.28 × 105

Tardigrade Admin · Accepted Answer

NH3+H2O leftharpoons NH4++OH- At equi. (0.100-1.34 × 10-3)M 1.34× 10-3 M 1.34 × 10-3 M = 0.09866 M Kb=([NH4+[OH-]]/[NH3])=(1.34×10-3×1.34×10-3/0.09866) =1.8199 × 10-5≈1.82 ×10-5

Q. When $0.100$ mole of ammonia, $N H_{3}$ is dissolved in sufficient water to make $1.0 L$ of solution, the solution is found to have a hydroxide ion concentration of $1.34 \times 1 0^{- 3} M$ . Calculate $K_{b}$ for ammonia.

$1.56 \times 1 0^{3}$

$1.82 \times 1 0^{- 5}$

$1.37 \times 1 0^{- 3}$

$1.28 \times 1 0^{5}$

Q. When 0.100 mole of ammonia, NH3​ is dissolved in sufficient water to make 1.0L of solution, the solution is found to have a hydroxide ion concentration of 1.34×10−3M. Calculate Kb​ for ammonia.

1.56×103

1.82×10−5

1.37×10−3

1.28×105

NH3​+H2​O⇌NH4+​+OH− At equi. (0.100−1.34×10−3)M1.34×10−3M1.34×10−3M=0.09866M Kb​=[NH3​][NH4+​[OH−]]​=0.098661.34×10−3×1.34×10−3​ =1.8199×10−5≈1.82×10−5

Q. When $0.100$ mole of ammonia, $N H_{3}$ is dissolved in sufficient water to make $1.0 L$ of solution, the solution is found to have a hydroxide ion concentration of $1.34 \times 1 0^{- 3} M$ . Calculate $K_{b}$ for ammonia.

$1.56 \times 1 0^{3}$

$1.82 \times 1 0^{- 5}$

$1.37 \times 1 0^{- 3}$

$1.28 \times 1 0^{5}$