Answer:
See explanation
Step-by-step explanation:
pOH = -log[OH^-]
pOH =-log[1×10^-4 molL-1]
pOH = 4
pH +pOH =14
pH = 14 -4
pH =11
b) pOH = -log[OH^-]
pOH =-log[4.9×10^-2molL-1]
pOH = 1.31
pH +pOH =14
pH = 14 -1.31
pH =12.69
c)
pH= -log[H^+]
[H^+] = Antilog [-pH]
[H^+] = Antilog [-3]
[H^+] = 1 × 10^-3 molL-1
d) pOH = -log[OH^-]
[OH^-] = Antilog[-pOH]
[OH^-] = Antilog[-10.4]
[OH^-] = 3.98 ×10^-11
[OH^-] [H^+] = 1 × 10^-14
[H^+] = 1 × 10^-14/3.98 ×10^-11
[H^+] =2.5 ×10^-4 MolL^-1
e) A solution of pH 9 is basic because the pH scale is calibrated in such as way that 0-6.9 is acidic, 7 is neutral and 8 - 14 is basic.
f)
[H^+] = number of moles/volume
[H^+] = 0.35 moles/8L
[H^+] = 0.044 MolL^-1
pH = - log [H^+]
pH = -log [0.044 MolL^-1]
pH= 1.36