The question above has already been answered by john. But I want to answer here a question which may come to ones mind, when one reads the above answer. Namely the following: The original question is concerned with the special case $C=W$ if one considers right calculus of fractions for a pair $(C,W)$. Calculus of right fractions says that,

1) spans can be completed to commutative squares, and

2) parallel arrows, which are coequalized by some arrow also get equalized by some other arrow.

Of course existence of pullbacks implies 1). But does 1) imply 2)? Here is a counterexample:

Define $C$ to be the category with objects natural numbers exluding $0$. An arrow from $n$ to $m$ is a sequence $(n_1,...,n_m)$ of $m$ natural numbers $\neq 0$ which sum up to $n$. Composition is as follows: Suppose you want to compose $(n_1,...,n_m):n\rightarrow m$ with $(m_1,...,m_k):m\rightarrow k$. To do this, divide the sequence $(n_1,...,n_m)$ into $k$ blocks of lengths $m_1,m_2,...,m_k$ and sum up the blocks, i.e. the composition is
$$(n_1+...+n_{m_1},n_{m_1+1}+...+n_{m_2},...)$$
Property 1) is not hard to verify. But property 2) is not satisfied since $(2,1)$ and $(1,2)$ are coequalized by $(2)$ but cannot be equalized.