Question

Calculate the ph of 0.002 N in NH4OH which undergoes 2.3% dissociation

Answer 1

Answer:

Explanation:

NH4OH→NH4⁺+ OH⁻

c=0.002*2.3*0.01=4.6×10⁻⁵=[OH⁻}

POH=-log[OH⁻]

POH=4.34

PH+POh=14

PH=9.66 ≅10