`^^_(m_((NH_(4)OH)))^(o)` is equal to

Options

A. `wedge_(m[NH_(4)OH])^(0)+wedge_(m[NH_(4)Cl])^(0)-wedge_([HCl])^(0)`
B. `wedge_(m[NH_(4)Cl])^(0)+wedge_(m[NaOH])^(0)-wedge_([NaCl])^(0)`
C. `wedge_(m[NH_(4)Cl])^(0)+wedge_(m[NaCl])^(0)-wedge_([NaOH])^(0)`
D. `wedge_(m[NaOH])^(0)+wedge_(m[NaCl])^(0)-wedge_([NH_(4)Cl])^(0)`

Answer

B

Explanation

`NH_(4)ClNH_(4)Cl hArr NH_(4)^(+)+Cl` . . . (i)
`NaCl hArr Na^(+)+Cl` . . . (ii)
`NaOH hArr Na+OH` . . . (iii)
`NH_(4)OH hArr NH_(4)^(+)+OH^(-)` . . . (iv)
To get equation (iv)
`wedge_(m)^(@)(NH_(4)Cl)+(NaOH)^(-)wedge_(m)^(@)(NaCl)=wedge_(m)^(@)(NH_(4)OH)`