sebanyak 500mL larutan NH4CL mempunyai ph=5. jika kb NH4OH = 10^-5 dan Mr NH4CL = 53.5. Massa garam NH4CL yang terlarut sebanyak..???

ph =5
[H+] = 10^-5
[H+] = √Kw/Kb * [G]
10^-5 = √10^-14/10^-5*[G]
(10^-5)^2= 10^-9*[G]
10^-10/10^-9=[G]
[G]= 10^-1

[G]= m/mr*1000/v
10^-1=gr/53.5*1000/500
m= 53.5*0.1/2
m= 2.675 gram